Lecture Notes

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Philosophy

PHIL10052: Introduction to Logic

Dr Graham Stevens

Lecture Notes and Tutorial Exercises 2010

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Note These lecture notes have been produced for first year logic students at Manchester since 2004. During that time they have been constantly updated and improved, often thanks to comments from students on the course. If you spot any typographical errors or ambiguities in the text, or have any general comments to make, please email them to me at the address below so that your comments can be taken on board for the benefit of future students on this course. The Philosophy Discipline Area awards the annual Polanyi Logic Prize for the best performance in the examination for this course. [email protected]

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Logic lecture 1

PL: Truth-functions and formation rules.

1. Propositions vs sentences In this course we are going to spend most of the time learning a fairly simple language. This ‘artificial language’ we will call PL (for ‘propositional logic’). Strictly speaking, we will not be studying a logic of propositions on this course, however. We will be studying the logic of a set of artificially introduced symbols which we will sometimes call (for want of a better name) ‘formulas’. Intuitively, however, we will usually interpret these formulas as representing certain kinds of sentences (the kind that express propositions). The kinds of sentences we will be concerned with are declarative sentences (sentences which are true or false). In fact, their truth or falsehood is the only feature of them we will be interested in. From the outset, we will need to make an assumption that every sentence of our language is either true or false (different languages of logic which reject this assumption are called ‘deviant’ or ‘non-classical’ logics). As we are only interested in the truth-values assigned to our sentences it really doesn’t matter to us what they mean. Therefore, instead of using English sentences like ‘Tony Blair drinks beer’, we will use letters to stand for sentences. We will use upper-case letters from the Roman alphabet ‘A, B, C, …Z’. These we will call, for obvious reasons, ‘sentence-letters’.1 So far this will be familiar to you from the Critical Thinking course. However, we will make things a lot more formal on this course. We will start by specifying the language PL in very precise terms. 2. Vocabulary and Grammar Our language consists of far fewer signs than English. We have our sentence-letters, signs for punctuation, and truth-functions. The truth-functions are the crucial part (hence they are sometimes referred to as the logical vocabulary). You have already met them in the Critical Thinking course and and we are going to spend the next few weeks getting to know how they work and what they mean in a far deeper way. For the time being, just a quick reminder will suffice: Truth-functions are sentential connectives. They connect sentences in such a way as to produce new sentences whose truth-values are determined by the truth-values of the sentences they connect. So for example, ‘It is not the case that snow is white’, ‘snow is white and grass is green’, ‘snow is white or grass is green’, ‘if snow is white then grass is green’, and ‘snow is white if and only if grass is green’ are examples of sentences formed in English by applying truth-functions (highlighted in bold) to the 1 You might think this means there are only 26 sentence letters in PL. You would be wrong: we can in principle get an enormous number by adding a number to any letter, e.g: ‘A1, A2, … An’. We will never actually use these but if you meet them in anything you read, just bear in mind that the use of these numbers is simply to extend the number of sentence letters beyond 26.

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sentences (highlighted in italics) ‘snow is white’ and ‘grass is green’. There are other possible truth-functions but these are the ones we will be interested in. In PL we will write them as follows:2 ‘~A’ means (roughly) ‘it is not the case that A’ or ‘not-A’. ‘(A ∧ B)’ means (roughly) ‘A and B’. ‘(A ∨ B)’ means (roughly) ‘A or B’. ‘(A ⊃ B)’ means (roughly) ‘If A then B’. ‘(A ≡ B)’ means (roughly) ‘A if and only if B’. We will often refer to the sentences of PL as formulas. The brackets are the only punctuation signs in PL. In the above formulas they are not really needed, but they become essential as our formulas become more complicated, as we shall soon see. Now that we have established our signs, we have to address how to use them. Just as in any other language, PL has a grammar. Grammar is far stricter in PL than in English. If you say something ungrammatical in English, we will usually be able to understand what you mean. In PL, an ungrammatical sentence is simply nonsense. We call ungrammatical sentences of PL ill-formed. We call grammatical sentences, well-formed formulas (wffs). A wff has to be constructed from our signs in accordance with formation rules. These are just rules of grammar for PL. They are as follows: An individual sentence letter is called an atomic formula. Any atomic formula is a wff. Attaching ‘~’ to any atomic formula yields a wff. So, e.g., ‘~A’, ‘~B’, ‘~Y’ are all wffs. Also ‘~’ can be attached to any wff to yield another wff. So, of course, ‘~~A’, ‘~ ~ ~A’, and ‘~ ~ ~ ~ ~ ~ ~A’ are all wffs. Let ‘φ’ and ‘ψ’ stand for any wffs of PL,3 then the following are all wffs: ~ φ (φ ∧ ψ) (φ ∨ ψ) (φ ⊃ ψ) (φ ≡ ψ) There are no other wffs.

2 These are the symbols used in Simpson’s book. Tomassi uses slightly different ones which are as follows: ‘&’ instead of ‘∧’, ‘→’ instead of ‘⊃’, and ‘↔’ instead of ‘≡’. ‘~’ and ‘∨’ are the same. This difference is of no consequence as the meanings are exactly the same. Whatever book you are using, if you are in any doubt about how to use elements of PL, ALWAYS FOLLOW THE USES SET OUT IN THE LECTURE NOTES. It will make life much, much easier. 3 ‘ϕ’ and ‘ψ’ are not part of the language PL. They are symbols that we are using to stand for the symbols of PL. In more technical terms they are symbols of a metalanguage (the language we use to talk about PL). This simply means that we only use them to explain things about PL: when we get on to arguing in PL we will not be allowed to use them (this is just the same as when we teach a language like French to English people—we can use English words to explain the grammar of French but, obviously, it is impossible to use English to speak in French!). We won’t be studying these kinds of things on this course but I mention it just because you may encounter the term in your reading (on any of your courses).

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Now that we have defined a wff we can see why brackets are so important. Take the following (ill-formed) formula: A ∨ B ⊃ C This is ambiguous: it could mean ‘A or if B then C’ or it could mean ‘if A or B then C’. Brackets are needed to distinguish them: ‘A ∨ (B ⊃ C)’ means ‘A or if B then C’. ‘(A ∨ B) ⊃ C’ means ‘if A or B then C’. Strictly speaking, we should have another pair of brackets to enclose the whole formula but as this leads to no confusion it makes life easier if we drop them. Every wff must have a main operator. In ‘A ∨ (B ⊃ C)’, ‘∨’ is the main operator. In ‘(A ∨ B) ⊃ C’, ‘⊃’ is the main operator. Correct use of brackets avoids confusion by making the main operator of every formula clear. If it is not made clear, the formula is not a wff, and not a sentence of PL. We have now learnt the very basics of PL. Don’t worry if it seems like a lot to digest right now. Like any language, it is only once you start using it that you really find yourself getting to grips with it. So far we have only learnt about the structure of PL and don’t yet know how to state things in it or construct arguments in it. We’ll spend plenty of time finding out how to use the language as the course goes on so be patient and persevere with it! Tutorial Exercises:

1. Why are the Greek letters ‘ϕ’ and ‘ψ’ used in our definitions of wffs not part of PL? 2. Why do we need brackets in PL?

(Exercises 3-12 are taken from Simpson, § 2.6): Which of the following are strictly speaking wffs (hint: remember that strictly speaking we have to have outside brackets when we don’t have an atomic formula, although ‘~’ can always be attached to any wff to yield a wff)?

3. ~ ~ ~ B 4. ((A) ⊃ B) 5. (F ∧ (G ∨ H)) 6. ((F ∧ G) ∨ H) 7. ((F ∧ G ∨ H)) 8. ~ (~ M ⊃ ((R ∨ G) ≡ S)) 9. ~ (~ M ⊃ (((R ∨ G) ≡ S)) 10. ((F ∨ G) ∧ (F ∨ G)) 11. R ∨ S 12. (f ⊃ (g ∨ s)) 13. Is it possible for the number of left-hand brackets in a wff to differ from the number of right-

hand brackets?

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Logic lecture 2

Truth Tables: Logical Equivalences and Arguments

In Critical Thinking, we found out how to use truth-tables to (a) represent the meanings of the logical constants in a precise manner, and (b) test any formula of PL to find it if it is a tautology, contradiction, or contingency. Now we are going to see what else truth-tables are good for. 1. Truth-table conventions Before we begin, we need to introduce a helpful convention. In Critical Thinking we never considered truth-tables complex enough to need this convention, but such a luxury can’t be maintained any longer. When constructing a truth-table, from now on, you should number each column of the table so as to keep track of which order its value should be worked out in. First count how many columns there are. Now number whatever columns can be worked out just from the values of the sentence letters listed across from the left (it is not necessary to copy the values across under the individual sentence letters unless you find it helpful). Number these beginning with the number of columns there are and working down (e.g. if there are 9 columns, proceed 9, 8, 7 …) until you can do no more. Next number the ones that can be worked out from these and continue your downwards numbering sequence until you can do no more. Now number the ones that can be worked out once the previous columns have been filled in, and so on. The final number added (1) will be the main operator. Here’s an example to demonstrate: A B C D (((~ A ∨ B) ⊃ (C ⊃ D)) ∧ (A ⊃ B)) ⊃ (A ⊃ (C ⊃ D)) T T T T F T T T T T T T T T T T F F T F F F T T F F T T F T F T T T T T T T T T T F F F T T T T T T T T T F T T F F T T F F T T T T F T F F F T F F F T F F T F F T F F T T F F T T T T F F F F F T T F F T T T F T T T T T T T T T T T T F T T F T T F F F T T T F F T F T T T T T T T T T T F T F F T T T T T T T T T F F T T T T T T T T T T T F F T F T T F F F T T T F F F F T T T T T T T T T T F F F F T T T T T T T T T 9 8 6 7 4 5 1 2 3 As the column under the main operator (marked ‘1’) contains only Ts, we have proved that this sentence is a tautology. Some things to note about this example are:

• The order in which each column is calculated is given in descending order by the numbers under the columns. So column 9 is done first, column 1 is the last (the main operator). It is important that the columns for least complex

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formulas are calculated first—e.g., in ‘~ A ∨ B’, the values for ‘~ A’ must be calculated before we move on to ‘∨’. For some formulas it doesn’t matter which is calculated first. For example, we could have calculated the values for the column marked ‘3’ before that marked ‘5’ in the above example. Obviously, however, we couldn’t have done column 2 before column 3 as we needed the result of 3 to calculate 2. Just remember – always calculate from the atomic formulas upwards, and ALWAYS NUMBER THE COLUMNS IN REVERSE ORDER.

• The sentence letters should always be listed on the left in alphabetical order. The result of the truth-table wouldn’t turn out any differently if we failed to do this but it is a convention which is used to make life easier.

• When you fill out the truth-possibilities under the sentence letters on the left-hand side of the truth-table, you should always follow the convention given in the above examples. Start with the letter on the right and write the possibilities as alternating Ts and Fs. In the next column to the left, alternate after every two Ts or Fs. In the next column alternate after every 4. In the next column after every eight, in the next column after every sixteen, and so-on. Again there are other ways of doing this but it is the universally accepted convention among logicians. Following it will make the process easier.

2. Testing for Logical Equivalence One of the most useful functions of a truth-table is to test whether two or more wffs are truth-functionally equivalent. Recall the truth-table for the triplebar (where, as usual, the Greek letters stand for any wff of PL): φ ψ φ ≡ ψ T T T T F F F T F F F T This states that two wffs are equivalent when and only when they are alike in truth-value. So a statement like ‘A ≡ B’ is true if and only if ‘A’ and ‘B’ are both true or ‘A’ and ‘B’ are both false. Whenever ‘A’ and ‘B’ differ in truth-value, ‘A ≡ B’ is false and we say that ‘A’ and ‘B’ are not truth-functionally equivalent. It is fairly simple, therefore, to see how we can use truth-tables to test whether wffs are truth-functionally equivalent. All we need to do is to draw out their truth-tables and compare them to see whether they always match in truth-value or not. Take, for example, ‘A ∨ B’ and ‘~ B ⊃ A’. These are truth-functionally equivalent. Let’s check. To do this, all we need to do is put the two wffs onto a truth-table as follows: A B A ∨ B ~ B ⊃ A T T T F T T F T T T F T T F T F F F T F 1 2 1

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Comparing the columns (in bold) under the main operators of each formula we can see that they are the same in truth-value. (Notice that, as we have two separate formulas here as opposed to one complex formula, we number each column of the truth-table for that formula and start the numbering again for the next formula). If we had wanted to, we could have gone about showing this by using the truth-table method to demonstrate that the formula ‘(A ∨ B) ≡ (~B ⊃ A)’ is a truth-functional truth of PL: A B (A ∨ B) ≡ (~ B ⊃ A) T T T T F T T F T T T T F T T T F T F F F T T F 2 1 4 3 In practice, however, it is easiest to demonstrate equivalences the first way. This is because we may sometimes want to show that more than two formulas are equivalent. In the following example, we show that ‘A ⊃ B’, ‘~ B ⊃ ~ A’, and ‘~A ∨ B’ are all truth-functionally equivalent: A B A ⊃ B ~ B ⊃ ~ A ~A ∨ B T T T F T F F T T F F T F F F F F T T F T T T T F F T T T T T T 1 3 1 2 2 1 Comparison of the values under the main operators shows again that these are truth-functional equivalences. 3. Using truth-tables to test the validity of arguments. If all has gone well, you ought to remember the definition of a valid argument given last semester. In case things have not gone well, or you are new to the course this semester, here it is again: Validity: An argument is valid if and only if, if the premises of an argument are

true, then the conclusion must also be true. Of course, this does not mean that a valid argument actually does have true premises: it only means that the conclusion would have to be true if the premises are true (if an argument is valid and its premises actually are true, it will be recalled, the argument is sound). Checking the validity of an argument in PL is a very simple business indeed (this is one of the reasons why PL is preferable to natural languages in some respects). The easiest way to do so is purely mechanical. We simply write a truth table much the same as the ones we have used for checking truth-functional equivalences but this

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time, instead of looking for our formulas to have the same truth-values in all possibilities, we are going to isolate one of the formulas as a conclusion and then check that this is never false when the premises are true. If it meets this condition it is valid. Before we do this, we will pause to introduce some helpful symbols and terminology that will make life easier in dealing with arguments in PL. A standard way of writing out arguments (used in Critical Thinking) is as follows: A A ⊃ B _____ B Here the two formulas above the line are our premises, and the formula below the line is the conclusion. The line, then, is the ‘inference bar’ and can be read as ‘therefore’. Simpson in fact uses this form to write arguments. We are going to adopt a more economical way of writing this and to do so, we will need a new symbol: ‘|=’. This symbol is called the turnstile (or, more officially, the semantic turnstile, but we don’t need to worry about that just yet).4 It replaces the inference bar so that we can now write the above argument as: A, A ⊃ B |= B This states that ‘B’ is a logical consequence of ‘A’ and ‘A ⊃ B’. You may prefer to read it as ‘A. If A then B. Therefore B.’ It basically just expresses the notion of validity or logical consequence that we are already familiar with in studying arguments. A few important things to note about the turnstile:

• The formulas (premises) on the left of the turnstile should be separated by commas so as to avoid any confusion.

• There is no limit on how many formulas can stand on the left of the turnstile. We are simply saying that what stands to the right follows from what stands on the left. So all of the premises of an argument go on the left and the conclusion stands on the right.

• In the case of a truth-functionally true formula, it follows from everything and nothing. We will therefore permit the turnstile to have nothing at all to its left, so long as we have a truth-functionally true formula on the right. E.g. ‘|= A ∨ ~A’. In such cases we can read the turnstile as ‘it is a logical truth that…’.

• What goes on the left of the turnstile is a set of formulas. When we are using the turnstile to write an argument, we are saying that the conclusion is a logical consequence of the set of premises. The argument (premises, turnstile and conclusion) taken as a whole is called a sequent.

• When we prove the validity of a sequent in PL, it should be noted that what we are doing is showing that a wff follows from a set of wffs in PL. For this reason, if we wanted to be perfectly clear about what we were proving we would have to show this by writing ‘|=PL’ instead of just ‘|=’ so as to show that

4 We will encounter a similar symbol in the next weeks: ‘|-’. This is a syntactic turnstile. I’ll explain the differences and relations between them when we get on to it.

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we mean the conclusion follows from the premises in PL. However, as we are mainly focusing on PL alone, we can omit this. If you encounter things like this in what you read, however, that is all it means.

• The turnstile is not part of PL; it is one of those symbols that we use to say things about formulas of PL.

Let’s now use a truth-table to prove a sequent. We will show that the sequent ‘A, A ⊃ B |= B’ is valid in PL: A B A A ⊃ B |= B T T T T T T F T F F F T F T T F F F T F If we look at the table we see there is only one row (the 1st one) in which all of the premises are true. Looking across to the conclusion on this row, we see that the conclusion is also true on this row. Therefore the argument is valid. Note that the 2nd and 4th rows have one true premise and a false conclusion. This does NOT make the argument invalid however. An argument is invalid only if it is possible for all of its premises to be true and its conclusion false. Hence we have proved our sequent. Here is an example of an invalid sequent:

A ⊃ B, ~A |= ~B. A B A ⊃ B ~A |= ~B T T T F F T F F F T F T T T F F F T T T Look at row 3 (marked in bold). Here we have all true premises and a false conclusion. Hence we have disproved the sequent.

It is worth noting that any sequent with a truth-functionally true formula to the right of the turnstile must be valid. An argument can only be invalid if the truth of all its premises lead to a false conclusion, but if the conclusion cannot be false, it is evident that the argument cannot be invalid.

3. Tutorial Exercises

1. What was the fallacy employed in the invalid sequent at the end of section 3? For each of the following pairs of formulas, construct a truth-table to determine whether they are truth-functionally equivalent.

2. (F ⊃ G) ∧ (F ⊃ H) F ⊃ (G ∧ H) 3. (F ∧ G) ⊃ H (F ⊃ H) ∧ (G ⊃ H) 4. M ⊃ N ~ N ⊃ (M ⊃ N)

For each of the following sequents, construct a truth-table to determine if it is truth-functionally valid.

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5. F ⊃ G, F |= F ∧ G 6. P ≡ ~ Q, Q ∨ P |= ~ P 7. (R ⊃ S) ∨ (S ⊃ N), ~ N ⊃ (R ∧ S) |= S ⊃ ~ R 8. ~ K ∧ V, K ∨ ~ M, T ⊃ ~ V |= M ∧ ~ T

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Logic lecture 3 Proof (I)

1. What is a proof? The main purpose of a language of formal logic like PL is to prove things. Only when you have learned to prove things in PL will you have really learned the language. Of course, we have already done this to some extent by using truth-tables but there are some important differences between truth-table demonstrations of truth-functional truth or validity and formal proofs (or derivations as they are sometimes called). There are two very important differences:

1. A truth-table demonstrates that a formula is a tautology or that an argument is valid by studying the truth-possibilities (or, as we sometimes say, interpretations) of formulas; a proof, however, demonstrates that a formula is a theorem purely by implementing rules for using the signs of PL.

2. A truth-table is a purely mechanical procedure which will tell us for any wff of PL, whether it is a tautology, a contradiction, or a contingent statement (and whether any argument is valid or invalid). In PL we can always do this but in other more sophisticated languages (such as QL which we will look at in a few weeks) it is not possible to prove things by a mechanical method like the truth-table method.5

Because a proof is not a mechanical procedure like a truth-table, it requires more thought to construct. This means that you might find proofs difficult compared to truth-tables. But it also means that you ought to find them far more satisfying to complete. In fact, proofs can come in various forms. The two main methods of proving things in logic are by the axiomatic technique and the natural deduction technique. The axiomatic method is the oldest, and natural deduction is a method invented to simplify things. PL is a system of natural deduction; we will not be looking at all at axiomatic proofs. A system of natural deduction proves things just by using rules which allow new lines to be added to a proof in order to try and reach the desired conclusion. A proof is simply a finite list of formulas which are arranged in a numbered list. Each line of the list is introduced in accordance with the rules that we are about to specify. Here is a very simple example of a proof: 1 A P 2 A ⊃ B P 3 B 1,2, ⊃E The ‘P’s written to side of lines 1 & 2 denote that these lines are premises. Line 3 then derives ‘B’ from ‘A’ and ‘A ⊃ B’ by a rule called ‘⊃E’ (horseshoe elimination). The 1 & the 2 next to ‘⊃E’ refer to the lines containing the formulas which justify the use of this rule. The line to the right of the numbers of each line is called the scope line.

5 The problem of finding this out was called the ‘decision problem’ and took enormous ingenuity to solve. The solution was eventually found (independently) by Alonzo Church and Alan Turing.

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Rules like ‘⊃E’ are called rules of inference or introduction and elimination rules. They are one of four ways in which a new line of a proof can be justified. The full list of ways to justify the introduction of a formula in a proof is:

1. The formula is a premise 2. The formula is an assumption 3. The formula is reiterated from an earlier stage of the proof 4. The formula is obtained by a rule of inference

All of these notions will be explained over the next few weeks. (1) is easily explained. Before doing so, we will introduce a new symbol. The symbol ‘|-’ is called the syntactic turnstile. Whereas ‘|=’ means ‘is a logical consequence of’, ‘|-’ means ‘is derivable from’. When we construct proofs we will often want to proof that sequents are valid in PL. In the case of truth-tables, ‘|=’ was the right relation as we were talking about wffs following from other wffs in accordance with their truth-possibilities. Now that we are talking just about proving validity by the use of rules, we use ‘|-’. 2. Some Rules of Inference Take the sequent ‘A, A ⊃B |- B’. In this sequent, everything to the left of ‘|-’ are premises. Therefore, as in the above example, we introduce these first of all into our proof and mark each one on the right with a ‘P’ to make plain that it is a premise. Our job is then to try and get from these premises to the conclusion ‘B’. To do this, we need to know the rules of inference. So let’s look at some. There are 10 main rules of inference in PL: 2 for each of the logical operators. In addition, we will also have rules for reiteration and for making assumptions. For the moment, however, we will just learn some of the simpler rules for the operators. The first one is very easy indeed: caret introduction or ‘∧I’6. Letting ‘φ’ and ‘ψ’ stand for any two wffs of PL: If we already have on earlier lines of a proof (on the same scope line) the 2 formulas ‘φ’ ‘ψ’, then we can introduce the formula ‘φ ∧ ψ’ on a new line (on the same scope line). The full definition is:

∧I ONE CAN WRITE THE FORMULA ‘φ ∧ ψ’ OR THE FORMULA ‘ψ ∧ φ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT BOTH:

1. THE FORMULA ‘φ’ IS ALREADY ON THE SAME SCOPE LINE. 2. THE FORMULA ‘ψ’ IS ALREADY ON THE SAME SCOPE LINE.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘φ’ OCCURS, THE NUMBER OF THE LINE WHERE ‘ψ’ OCCURS AND THE RULE, ‘∧I’.

An example using this rule is the following proof of ‘A, B |- A ∧ B’

6 Strictly speaking we should always use the name of the symbol when talking about the rule for introducing it (so, e.g. ‘caret introduction’ rather than ‘conjunction introduction’ or ‘and introduction’). Speaking loosely, however, there is no real harm in using the more informal names if you prefer.

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1 A P 2 B P 3 A ∧ B 1,2, ∧I Note also that ‘φ’ and ‘ψ’ do not have to be distinct. The following use of ∧I is legitimate: 1 A P 2 A ∧ A 1, ∧I The next rule is just as simple. If we have any conjunction ‘φ ∧ ψ’ then we can infer either ‘φ’ or ‘ψ’ (or both). This is caret elimination:

∧E ONE CAN WRITE THE FORMULA ‘φ’ OR THE FORMULA ‘ψ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT: THE FORMULA ‘φ ∧ ψ’ IS ALREADY ON THE SAME SCOPE LINE. THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘φ ∧ ψ’ OCCURS AND THE RULE, ‘∧E’.

An example is the following proof of ‘A ∧ (B ∧ C) |- C’: 1 A ∧ (B ∧ C) P 2 B ∧ C 1, ∧E 3 C 2, ∧E (Notice that we had to eliminate the main operator in line 1 before we could re-apply ∧E to line 2 in order to get ‘C’). The next rule is tilde elimination. From ‘~ ~ φ’ one can infer ‘φ’:

~E ONE CAN WRITE THE FORMULA ‘φ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT: THE FORMULA ~ ~ φ IS ALREADY ON THE SAME SCOPE LINE. THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘~ ~ φ’ OCCURS AND THE RULE ‘~E’.

It is important to remember again that this rule can only be used on a formula whose main operator is a ‘~’ followed by another ‘~’ (it couldn’t be applied to ‘~ ~ A’ in ‘B ∧ ~ ~ A’ until ∧E had been used to get ‘~ ~ A’, for example.

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Our next rule is the rule of vel introduction. For any formula ‘φ’, we can always infer ‘φ ∨ ψ’:

∨I ONE CAN WRITE THE FORMULA ‘φ ∨ ψ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT EITHER ONE OF THE FOLLOWING CONDITIONS IS MET:

1. THE FORMULA ‘φ’ IS ALREADY ON THE SAME SCOPE LINE. 2. THE FORMULA ‘ψ’ IS ALREADY ON THE SAME SCOPE LINE.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘φ’ OR ‘ψ’ OCCURS AND THE RULE, ∨I.

Here is an example which uses ∨I several times: 1 A P 2 A ∨ B 1, ∨I 3 (A ∨ B) ∨ C 2, ∨I 4 ((A ∨ B) ∨ C) ∨ D 3, ∨I Our next rule is called horseshoe elimination. It is, again intuitively obvious. If we have ‘φ’ and ‘φ ⊃ ψ’, we can infer ‘ψ’:

⊃E ONE CAN WRITE THE FORMULA ‘ψ’ ON ANY GIVEN SCOPE LINE PROVIDED THAT BOTH:

1. THE FORMULA ‘φ’ IS ALREADY ON THE SAME SCOPE LINE. 2. THE FORMULA ‘φ ⊃ ψ’ IS ALREADY ON THE SAME SCOPE LINE.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘φ’ OCCURS, THE NUMBER OF THE LINE WHERE ‘φ ⊃ ψ’ OCCURS AND THE RULE, ‘⊃E’.

For example: 1 A P 2 B P 3 A ⊃ (B ⊃ C) P 4 B ⊃ C 1,3, ⊃E 5 C 2,4, ⊃E The last rule that we shall consider this week is the rule of triplebar elimination. This rule intuitively captures the equivalence between ‘A ≡ B’ and ‘(A ⊃ B) ∧ (B ⊃ A)’:

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≡E ONE CAN WRITE THE FORMULA ‘ψ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT BOTH OF THE FOLLOWING CONDITIONS ARE MET:

1. THE FORMULA ‘φ’ IS ALREADY ON THE SAME SCOPE LINE. 2. THE FORMULA ‘φ ≡ ψ’ IS ALREADY ON THE SAME SCOPE LINE.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘φ’ OCCURS, THE NUMBER OF THE LINE WHERE ‘φ ≡ ψ’ OCCURS, AND THE RULE ‘≡E’. SIMILARLY, ONE CAN WRITE THE FORMULA ‘φ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT BOTH OF THE FOLLOWING CONDITIONS ARE MET:

1. THE FORMULA ‘ψ’ IS ALREADY ON THE SAME SCOPE LINE. 2. THE FORMULA ‘φ ≡ ψ’ IS ALREADY ON THE SAME SCOPE LINE.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘ψ’ OCCURS, THE NUMBER OF THE LINE WHERE ‘φ ≡ ψ’ OCCURS, AND THE RULE ‘≡E’.

Here are some examples of the correct use of ≡E: 1 A P 2 B P 3 C ≡ A P 4 C ≡ (B ≡ D) P 5 C 1,3, ≡E 6 B ≡ D 4,5, ≡E 7 D 2,6, ≡E

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Tutorial Exercises (Taken from Simpson § 3.8): Each of the following proofs has several justifications missing. Supply the justifications, giving the name of the rule and the line numbers which justify the use of the rule.

Question 1 1 A P 2 ~ ~ G P 3 G 4 A ∧ G 5 (A ∧ G) ∨ ~ G

Question 2 1 R ≡ S P 2 Q P 3 (Q ∨ P) ⊃ (M ∧ N) P 4 N ⊃ S P 5 Q ∨ P 6 M ∧ N 7 N 8 S 9 R

Question 3 1 A P 2 R P 3 A ⊃ (R ≡ B) P 4 B ⊃ ((R ⊃ Q) ∧ (B ⊃ C)) P 5 R ≡ B 6 B 7 (R ⊃ Q) ∧ (B ⊃ C) 8 R ⊃ Q 9 B ⊃ C 10 Q 11 C 12 Q ∧ C 13 (Q ∧ C) ∨ (F ∧ L)

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Logic lecture 4

Proof (II): Strategies and Assumptions

1. How do you construct a proof? We looked last week at some of the rules of inference (introduction and elimination rules) for proofs in PL. Before we go on to look at some more, we should pause to think about what to do with these rules: how do we go about proving things with them? We already know that this is going to be a step-by-step process of moving by one rule at a time from one line of proof to another. But what we don’t want to to do is to find ourselves just wildly introducing rule after rule without getting anywhere near our desired conclusion. What we need, in other words, is a strategy. Let’s suppose we want to prove the following sequent: A ∧ B, (C ∨ D) ⊃ G, B ⊃ C |- G What strategy should we adopt for proving it? Well, let’s start with what we know. We know what our premises are and what our desired conclusion is: 1 A ∧ B P 2 (C ∨ D) ⊃ G P 3 B ⊃ C P z G ? We don’t know how many steps it will take to get to our conclusion, so we put a ‘z’ for the time being, letting ‘z’ just stand for whatever number it turns out to be. We put a ‘?’ next to it to show that we need to find the way to get this line. Let’s call this kind of half-completed proof a proof-sketch. Our question, then, is how do we get from 1, 2, & 3 to z? Well, looking at line 2 we can see that if we had ‘(C ∨ D)’, this will give us ‘G’ by ⊃E. So let’s try and get this. We now update our proof-sketch with our new target, also keeping track on the right-hand side of how that will give us the line we want: 1 A ∧ B P 2 (C ∨ D) ⊃ G P 3 B ⊃ C P y (C ∨ D) ? z G 2,y, ⊃E We don’t yet know how to get line y, however. So let’s re-apply the same line of reasoning here. We know that we could get ‘(C ∨ D)’ by ∨I if we had ‘C’ or ‘D’.

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There is no obvious way of getting ‘D’, but we can get ‘C’ from line 3 if we have B by the rule of ⊃E. This is promising; let’s pause to update our proof-sketch: 1 A ∧ B P 2 (C ∨ D) ⊃ G P 3 B ⊃ C P w C 3,?, ⊃E y (C ∨ D) w, ∨I z G 2,y, ⊃E So now we have a new target: we need to obtain ‘B’ from somewhere. This is easy enough, as we can just apply ∧E to line 1. This completes the proof: 1 A ∧ B P 2 (C ∨ D) ⊃ G P 3 B ⊃ C P 4 B 1, ∧E 5 C 3,4, ⊃E 6 (C ∨ D) 5, ∨I 7 G 2,6, ⊃E Once we have used our proof-sketch to find all of the steps, we can write up the proof in its final form, replacing our ‘x’ ‘y’ and ‘z’s with the correct numbers. Sometimes, particularly once you have got good at constructing proofs, you may not need to sketch the proof out in this way. Usually, however, a proof will be sufficiently complicated that you will need to sketch the proof. When you sketch a proof, no matter how complicated it is, you should apply the strategy just outlined and most importantly of all, you should always think one step at a time. When you sketch a proof, keep your rough sketch until you are sure you have got the proof right, then write it up afresh neatly, making sure every line is justified by the rules of PL. Remember that anything which is not fully justified by the rules of PL does not count as a proof in PL at all. 2. Assumptions One of the things we do, in our everyday reasoning, is offer arguments in support of conditional statements. We say, ‘if you pass logic, then you won’t have to resit it,’ and then offer justifications for the claim. A standard, and very effective, way of doing so is to assume the antecedent: we say ‘assume that you do pass logic, then …’. Notice that the sentence, ‘if you pass logic, then you won’t have to resit it’, is true regardless of whether you actually do pass logic or not: if you fail logic, thus having to do a resit, it doesn’t stop the sentence ‘if you pass logic, then you won’t have to resit it’ from being true. Similarly, in PL, we are often going to want to establish conditional claims. This kind of process is sometimes called ‘conditional proof’. We need to get a hold on this notion before we can progress to our remaining rules of inference.

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In a proof of PL, our way saying, ‘assume that A’, is to introduce ‘A’ on a new scope line. We call this the secondary scope line. It works in pretty much the same way as our ordinary habit of saying ‘assume that…, then it follows that…’. The following proof shows an example of how to employ a secondary scope line. In the proof, we also use two new rules, called ‘R’ and ‘⊃I’. Don’t worry about these for the moment, we’ll get on to them shortly. 1 A ∧ B P 2 Q A 3 A ∧ B 1, R 4 A 3, ∧E 5 Q ⊃ A 2-4, ⊃I The secondary scope line begins at line 2 and ‘Q’ is the assumption. When an assumption is introduced, we always place it on a new scope line, and give it the justification ‘A’ (assumption) on the right. A new scope line always begins with an assumption and an assumption can only ever be introduced at the beginning of a new scope line. There is no limit to how many new scope lines we can introduce, so long as we follow the rules correctly. For example: 1 A P 2 B A 3 C A 4 A 1,R 5 C ⊃ A 3-4, ⊃I 6 B ⊃ (C ⊃ A) 2-5, ⊃I Each of the secondary scope lines marks extent of a sub-proof or sub-derivation. When a secondary scope line is completed, we use a rule of inference to finish the line and return to the next scope line to the left, taking with us the information we have picked up on the secondary scope line. Let’s now look at some of those rules, and state the rule for introducing assumptions more precisely. The first two rules to deal with are our rule of assumption and the rule called ‘R’ that has cropped up in the previous examples.

A AT ANY POINT IN A PROOF, ONE CAN INTRODUCE ANY WFF AS AN ASSUMPTION, PROVIDED THAT: THE FORMULA IS WRITTEN AT THE BEGINNING OF A NEW SCOPE LINE. THE FORMULA WHICH IS ASSUMED SHOULD BE FOLLOWED BY A SHORT HORIZONTAL LINE TO DISTINGUISH IT FROM OTHER FORMULAS WHICH MAY APPEAR ON THE NEW SCOPE LINE. THE JUSTIFICATION FOR THE NEW LINE IS JUST THE RULE ‘A’.

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The rule of reiteration is very simple to understand. If we know that ‘A’ is true, then whatever else we find out, we will still know that ‘A’ is true. So we can always draw on things we already know to use on our secondary scope lines. More formally:

R AT ANY POINT ON A SCOPE LINE, ONE CAN COPY ANY FORMULA WHICH ALREADY APPEARS IN THE PROOF ONTO THE SCOPE LINE, PROVIDED THAT: THE SCOPE LINE FROM WHICH THE FORMULA IS COPIED HAS NOT BEEN ENDED. THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE FROM WHICH THE FORMULA IS REITERATED AND THE RULE ‘R’.

Let’s go straight on to the other rule that came up in the proofs we have looked at: the rule of horsehoe introduction. If we have assumed A, and shown that B follows from A, then we have shown that if A is true then B is true. This is reflected in the rule for introducing the horsehoe:

⊃I ONE CAN END ANY SUB-PROOF AND WRITE THE FORMULA ‘φ ⊃ ψ’ ON THE SCOPE LINE IMMEDIATELY TO THE LEFT OF THE ENDED SUB-PROOF, PROVIDED THAT BOTH OF THE FOLLOWING CONDITIONS ARE MET:

1. THE SUB-PROOF BEGINS WITH THE ASSUMPTION ‘φ’. 2. The formula ‘ψ’ appears in the sub-proof.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE RANGE OF NUMBERS IN THE ENDED SUB-PROOF OF ‘ψ’ FROM ‘φ’ AND THE RULE, ⊃I.

Examples of ⊃I can be found in our earlier proofs above. The rule for triplebar introduction is very much the same as 2 application of ⊃I. This is unsurprising: ‘φ ≡ ψ’ is truth-functionally equivalent to ‘(φ ⊃ ψ) ∧ (ψ ⊃ φ)’:

≡I ONE CAN WRITE THE FORMULA ‘(φ ≡ ψ)’ OR THE FORMULA ‘(ψ≡ φ)’ ON ANY SCOPE LINE, PROVIDED THAT BOTH OF THE FOLLOWING CONDITIONS ARE MET:

1. ON A SCOPE LINE IMMEDIATELY TO THE RIGHT OF THE GIVEN SCOPE LINE, THERE IS A SUB-PROOF OF ‘ψ’ FROM ‘φ’.

2. ON A SCOPE LINE IMMEDIATELY TO THE RIGHT OF THE GIVEN SCOPE LINE, THERE IS A SUB-PROOF OF ‘φ’ FROM ‘ψ’.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE RANGE OF NUMBERS OF THE SUB-PROOF OF ‘ψ’ FROM ‘φ’, THE RANGE OF NUMBERS OF THE SUB-PROOF OF ‘φ’ FROM ‘ψ’, AND THE RULE, ≡I

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Here are a couple examples of the correct use of ≡I. 1 (A ⊃ B) ∧ (B ⊃ A) P 2 A ⊃ B 1, ∧E 3 B ⊃ A 1, ∧E 4 A A 5 A ⊃ B 2, R 6 B 4,5, ⊃E 7 B A 8 B ⊃ A 3, R 9 A 7,8, ⊃E 10 A ≡ B 4-6,7-9, ≡I 11 B ≡ A 4-6,7-9, ≡I Finally, let’s have a proof that uses all of our new skills together. We will prove the sequent: (A ⊃ B) ∧ (B ⊃ A), C ⊃ F, F ⊃ D |- (A ≡ B) ∧ (C ⊃ D) 1 (A ⊃ B) ∧ (B ⊃ A) P 2 C ⊃ F P 3 F ⊃ D P 4 A ⊃ B 1, ∧E 5 B ⊃ A 1, ∧E 6 A A 7 A ⊃ B 4, R 8 B 6,7, ⊃E 9 B A 10 B ⊃ A 5, R 11 A 9,10, ⊃E 12 A ≡ B 6-8,9-11, ≡I 13 C A 14 C ⊃ F 2, R 15 F 13,14, ⊃E 16 F ⊃ D 3, R 17 D 15,16, ⊃E 18 C ⊃ D 13-17, ⊃I 19 (A ≡ B) ∧ (C ⊃ D) 12,18, ∧I 3. Tutorial Exercises Construct proofs to prove the validity of the following sequents (n.b. these do not require you to introduce any assumptions):

1. B, B ≡ ~ ~ A, (A ∧ B) ⊃ R |- R

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2. B, B ≡ ~ ~ A, (A ∨ S) ⊃ R |- R 3. A, A ⊃ G, B ≡ R, B ⊃ S, G ≡ ~ ~ I, ((F ∨ G) ∧ (H ∨ I)) ⊃ R |- A ∧ S

Construct proofs to prove the validity of the following sequents (n.b. these will require you to introduce assumptions):

1. F |- D ⊃ (E ⊃ F) 2. J ⊃ (K ⊃ L), J ⊃ K |- J ⊃ L 3. A ⊃ B, B ⊃ C, C ⊃ A |- A ≡ C 4. (J ∨ K) ⊃ L |- J ⊃ L

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Logic lecture 5

Proof (III): More Rules and Categorical Proofs 1. ~I and ∨E Each of the truth-functions of PL has 2 rules that govern its use in a proof: an introduction rule and an elimination rule. That means we have 2 rules left to learn: the introduction rule for negation and the elimination rule for disjunction. Both rules are a little tricky, but immensely useful once mastered. The rule of ~I, in particular, is very commonly used by mathematicians and (in a more informal way) by philosophers and is sometimes given other names like proof by contradiction or reductio ad absurdum. The essence of ~I is that, if we assume a formula ‘φ’ and then show that the assumption of ‘φ’ leads to a contradiction, we will have thereby demonstrated that ‘φ’ cannot be true. Hence we will be justified in asserting ‘~ φ’. Here is the rule set out more formally:

~I

ONE CAN WRITE THE FORMULA ‘~ φ’ ON ANY SCOPE LINE, PROVIDED THAT BOTH OF THE FOLLOWING CONDITIONS ARE MET:

1. ON A SCOPE LINE IMMEDIATELY TO THE RIGHT OF THE GIVEN SCOPE LINE, THERE IS A SUB-PROOF WHICH BEGINS WITH THE ASSUMPTION ‘φ’.

2. ON THE SCOPE LINE WHICH BEGINS WITH THE ASSUMPTION ‘φ’ THERE APPEAR TWO FORMULAS WHICH FORM A CONTRADICTION ‘ψ’ AND ‘~ ψ’.

THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE THE ASSUMPTION ‘φ’ IS INTRODUCED, THE NUMBER OF THE LINE WHERE ‘ψ’ OCCURS, THE NUMBER OF THE LINE WHERE ‘~ ψ’ OCCURS, AND THE RULE ~ I.

The following proofs demonstrate some uses of ~ I. 1 ~ (A ∨ B) P 2 C ∧ B A 3 B 2, ∧E 4 A ∨ B 3, ∨I 5 ~ (A ∨ B) 1, R 6 ~ (C ∧ B) 2,4,5, ~I

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1 A ∧ B P 2 ~ (A ∨ B) A 3 A ∧ B 1, R 4 A 3, ∧E 5 A ∨ B 4, ∨I 6 ~ ~ (A ∨ B) 2,2,5, ~I 7 A ∨ B 6, ~E Notice that in the last proof, the assumption itself was taken as one of the contradictory pairs. This is perfectly legitimate but the justification for the use of ~I should make 2 references to the assumption in such cases (see line 6). Another thing to note about the above demonstrations is that we can only negate the whole formula of the assumption if we show a contradiction follows from it; we cannot negate any part of the formula smaller than the whole. The rule of ∨E is perhaps the most difficult of all of the rules to master, but it is actually quite intuitive if you think it through carefully. Suppose that Rovers are playing United in the cup final and that both teams have French managers. Well you know that Rovers or United must win, hence you know that a team with a French manager will win the cup. Whenever we have a disjunction ‘φ ∨ ψ’ and we can prove that a formula ‘χ’ follows from both ‘φ’ and ‘ψ’, the rule of ∨E allows us to infer ‘χ’ (it is, of course, possible that ‘χ’ might actually be one of ‘φ’ or ‘ψ’, but it doesn’t by any means have to be).

∨E ONE CAN WRITE THE FORMULA ‘χ’ ON ANY SCOPE LINE, PROVIDED THAT ALL OF THE FOLLOWING CONDITIONS ARE MET:

1. THERE IS A DISJUNCTION ‘φ ∨ ψ’ ON THE GIVEN SCOPE LINE. 2. ON A SCOPE LINE IMMEDIATELY TO THE RIGHT OF THE GIVEN SCOPE

LINE THERE IS A SUB-PROOF OF THE FORMULA ‘χ’ FROM ‘φ’. 3. ON A SCOPE LINE IMMEDIATELY TO THE RIGHT OF THE GIVEN SCOPE

LINE THERE IS A SUB-PROOF OF THE FORMULA ‘χ’ FROM ‘ψ’. THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE ON WHICH THE DISJUNCTION ‘φ ∨ ψ’ APPEARS, THE TWO RANGES OF THE TWO SUB-PROOFS, AND THE RULE ∨E.

Here are a couple of examples:

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1 A ∨ B P 2 ~ A P 3 |B A 4 A A 5 ~ B A 6 ~ A 2, R 7 A 4, R 8 ~ ~ B 5,6,7, ~I 9 B 8, ~E 10 B 1,3-3,4-9, ∨E 1 (A ∨ B) ∨ (C ∧ D) P 2 (A ∨ B) ⊃ G P 3 (C ∧ D) ⊃ G P 4 (A ∨ B) A 5 (A ∨ B) ⊃ G 2, R 6 G 4,5, ⊃E 7 (C ∧ D) A 8 (C ∧ D) ⊃ G 3, R 9 G 7,8, ⊃E 10 G 1,4-6,7-9, ∨E The first proof is a very useful one as it shows us how to get one disjunct from the the disjunction if we can can prove that it follows from the other. Notice that, in this case, merely the assumption of the disjunct we are after (in this case ‘B’) counts as proof that it is derivable from that assumption as anything is obviously derivable from itself (hence our justification contains ‘3-3’ in the first proof). 2. Indirect Proofs and Categorical Proofs An indirect proof is a proof which assumes the negation of the desired formula and then establishes that this leads to a contradiction, allowing, by the rule of ~I, to introduce the negation of this formula (thus obtaining the negation of the negation of the desired formula). An application of ~E then yields the desired formula. An example was given in the second illustration of ~I above. The usefulness of indirect proof will become clear shortly.

So far we have been proving that sequents, or arguments, are valid in PL. But we also want to be able to prove that something is a theorem in PL in just the same way that we proved that some formulas were truth-functionally true by using truth-tables. In fact, proofs are far quicker than truth-tables for testing whether a formula is a theorem once the formula in question gets complicated. When we were working with truth-tables, we were working with the semantic turnstile ‘|=’. This, you will recall, can be read as ‘therefore’. If a formula of PL is truth-functionally true, then it can be proved without any premises, hence we can express this by simply putting a semantic turnstil in front of it with no premises before the turnstile. For example: ‘|= A ⊃ (A ∨ B)’ states that the formula ‘A ⊃ (A ∨ B)’ is truth-functionally true. This can be most naturally read as, “‘A ⊃ (A ∨ B)’ is a logical truth”. Similarly, we can use the

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syntactic turnstile ‘|-’, in ‘|- A ⊃ (A ∨ B)’ to say “‘A ⊃ (A ∨ B)’ is provable’ or “‘A ⊃ (A ∨ B)’ is a theorem.7 To prove this, we will need to show that the formula can be derived from no premises whatsoever. This kind of proof is called a categorical proof. As we have no premises at the start of a categorical proof, we have nothing to apply elimination rules to, so we have to start by using an assumption. Let’s take the example just discussed as a demonstration: ‘|-A ⊃ (A ∨ B)’. 1 A A 2 A ∨ B 1, ∨I 3 A ⊃ (A ∨ B) 1-2, ⊃I Now we can see why indirect proofs are so useful. Suppose we want to prove a logical truth like ‘|- A ∨ ~ A’. We can do so easily by indirect proof: 1 ~ (A ∨ ~ A) A 2 A A 3 A ∨ ~ A 2, ∨I 4 ~ (A ∨ ~ A) 1, R 5 ~ A 2,3,4, ~I 6 A ∨ ~ A 5, ∨I 7 ~ ~ (A ∨ ~ A) 1,1,6, ~I 8 A ∨ ~ A 7, ~E Categorical proofs are also very useful for proving truth-functional equivalences, as in the following proof of ‘|- (A ⊃ B) ≡ (~ A ∨ B)’. The proof, which is quite complicated, also demonstrates the usefulness of ∨E and ~I, and shows how sub-proofs can be used in an extended way to obtain the desired line of a proof when necessary. 7 We also spoke of “proving” things by using truth-tables. Strictly speaking, there is an important difference between showing that a formula is truth-functionally true (is a logical truth) and showing that it is a theorem (can be proved in a system—in this case PL). In the case of PL, however, we can actually prove that any logical truth is also a theorem and vice versa, that is to say, for any formula ‘φ’ of PL, we can prove ‘|=PL φ if and only if |-PL φ’. This kind of proof (called a metalogical proof) will not be covered on this course, however. In other, more complicated, languages of logic this close relation between logical truth and provability is not guaranteed but in PL we can always safely assume that the semantic and syntactic turnstiles coincide.

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1 A ⊃ B A 2 ~ (~ A ∨ B) A 3 A A 4 A ⊃ B 1, R 5 B 3,4, ⊃E 6 ~ A ∨ B 5, ∨I 7 ~ (~ A ∨ B) 2, R 8 ~ A 3,6,7, ~I 9 ~ A ∨ B 8, ∨I 10 ~ ~ (~ A ∨ B) 2,2,9, ~I 11 ~ A ∨ B 10, ~E 12 ~ A ∨ B A 13 ~ A A 14 ~ (A ⊃ B) A 15 A A 16 ~ A ∨ B 12, R 17 | B A 18 ~A A 19 ~ B A 20 A 15, R 21 ~A 18, R 22 ~ ~ B 19,20,21, ~I 23 B 22, ~E 24 B 16,17-17,18-23, ∨E 25 A ⊃ B 15-24, ⊃I 26 ~ ~ (A ⊃ B) 14,14,25, ~I 27 A ⊃ B 26, ~E 28 B A 29 A A 30 B 28, R 31 A ⊃ B 29-30, ⊃I 32 A ⊃ B 12,13-27,28-31, ∨E 33 (A ⊃ B) ≡ (~A ∨ B) 1-11, 12-32, ≡I

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This completes our survey of the rules for constructing proofs in PL. The following table summarises all of the introduction and elimination rules (the lower case letters on the left stand for line numbers): I E ∧ m φ m φ ∧ ψ m φ ∧ ψ n ψ or o φ ∧ ψ m,n, ∧I o φ m, ∧E o ψ m, ∧E ~ m φ A m ~ ~ φ n ψ n φ m, ~E o ~ ψ p ~ φ m,n,o,~I ⊃ m φ A m φ ⊃ ψ n ψ n φ o φ ⊃ ψ m-n, ⊃I o ψ m,n, ⊃E ∨ m φ m φ ∨ ψ n φ ∨ ψ m, ∨I n φ A

or o χ m φ p ψ A n ψ ∨ φ m, ∨I q χ r χ m,n-o,p-q, ∨E ≡ m φ A m φ ≡ ψ n ψ n φ o ψ A o ψ m,n, ≡E p φ or m φ ≡ ψ q φ ≡ ψ m-n,o-p, ≡I n ψ p φ m,n, ≡E

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3. A brief word on strategy With all of the rules in place, it is more important than ever to use them sensibly. If you introduce and eliminate without thinking about what you are aiming to achieve by it, any proof you hit upon will be a matter of luck, not skill. To keep things simple, remember what you learnt about strategy last week: always work one step at a time and work in rough, building upwards from your conclusion to see what you need to derive in order to reach that conclusion. When applying the rules to find a formula, you should follow this basic strategy:

1. See if the formula can be obtained by Reiteration; 2. See if the formula can be obtained by an Elimination Rule; 3. See if the formula can be obtained by an Introduction Rule; 4. See if the formula can be obtained by an indirect proof.

Tutorial Exercises Prove the following sequents in PL. 1. A ⊃ (B ∧ C), ~ C |- ~ A (hint: use ~I) 2. K ⊃ L |- (M ⊃ K) ⊃ (M ⊃ L) (hint: this proof relies heavily on ⊃I)8 3. A ≡ B, A ∨ B |- A ∧ B (hint: use ∨E) 4. F ⊃ ~ (G ∨ H), G |- ~ F (hint: use ~I) 5. (C ⊃ D) ∨ (C ⊃ E), C |- D ∨ E (hint: use ∨E) 6. |- ~ (P ∧ ~ P) (hint: categorical proof) 7. |- (P ∧ (P ⊃ Q)) ⊃ Q (hint: categorical proof)

8 As a rule of thumb, it is usually easier to prove a conclusion that is conditional in form by using ⊃I.

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Logic Lecture 6

QL

1. The Language QL We had a brief introduction to predicate logic in Critical Thinking (if you need a refresher revisit your notes). Let’s now set out a formal language QL based on the notions we were introduced to.9 We will do so in the same way we did for PL in lecture 1: we specify its vocabulary and its grammar in a precise way. Vocabulary:

The symbols of QL are: o Lower case letters from the beginning of the Roman alphabet a, b, c,

… are constants (names of individuals). o Upper case letters from the Roman alphabet A, B, C, … Z are n-placed

predicates (where n ≥ 1). o Lower case letters from the Roman alphabet x, y, z are variables. o ∀ is the universal quantifier. o ∃ is the existential quantifier. o ‘~’, ‘∧’, ‘∨’, ‘⊃’, ‘≡’ are sentential connectives. o Brackets ‘(‘ and ‘)’ are used for punctuation. o There are no other symbols.

Grammar: Let ‘φ’ stand for any n-placed predicate of QL, and let ‘α1 … αn’ stand for any n constants or variables of QL (where n ≥ 1). The following are wffs of QL:

φα1 … αn ~ φα1 … αn We call these formulas atomic formulas.

Let ‘ψ’ and ‘χ’ stand for any wff of QL. The following are also wffs of QL:

~ ψ (ψ ∧ χ) (ψ ∨ χ) (ψ ⊃ χ) (ψ ≡ χ)

9 The modern systems of predicate logic are all, essentially, forms of the language invented by Frege in his Begriffsschrift of 1879. Many of his ideas were discovered independently (though less clearly) by Pierce, Schröder, and Peano. If you are interested in such historical details, the best book is probably Kneale, W. & Kneale, M., The Development of Logic, Oxford University Press, 1962.

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Finally, let ‘ε’ stand for any variable of QL, and let ‘φ’ stand for any formula of QL containing at least one free10 occurrence of ε. Then the following are wffs of QL: (∀ε) φ

(∃ε) φ. Nothing is a wff of QL unless it can be derived from one or more application

of the preceding rules of grammar. Now let’s examine this formal system a bit more informally.

2. One-Placed Predicates and Many-Placed Predicates In any sentence, we can pick out a subject (or subjects) (the thing(s) the sentence is about) and a predicate (what is said about the things). We want our language to express this. We use lower case letters from the Roman alphabet (‘a’, ‘b’, ‘c’, … ‘z’) as names of individuals. We call these names constants. We use upper-case letters from the Roman alphabet (‘A’, ‘B’, ‘C’, … ‘Z’) as the predicates of QL.11 An individual is an actual thing—whatever is named by a constant (it might be a person, or a galaxy, or whatever else the sentence in question is about). A simple sentence like ‘Socrates is mortal’ can now be translated into QL. First, we isolate the predicate (in this case, ‘…is mortal’), then the individual (in this case Socrates). Now all we have to do is to translate the English predicate into QL (we’ll use ‘M’), and give Socrates a name (constant) in QL (we’ll use ‘s’): Socrates is mortal s: Socrates M…: …is mortal The dots after the predicate indicate the space in which a constant must go to yield a wff of QL (just as they do in ‘…is mortal’). So, we can now write our translation: Socrates is mortal s: Socrates M…: …is mortal Ms ‘Ms’ is a wff of QL. The sentence connectives are the same as those in PL. So we can now write complex sentences in QL. Suppose we wanted to say ‘Socrates is mortal only if Plato is a rascal’:

Socrates is mortal only if Plato is a rascal s: Socrates p: Plato M…: …is mortal R…: …is a rascal Ms ⊃ Rp 10 Free variables are variables which are not bound by a quantifier. So, e.g. ‘x’ is free in ‘Fx’ and in ‘(∀y) (Gy ⊃ Fx)’, but is bound in ‘(∀x) Fx’ and in ‘(∀y) (∃x) (Gy ⊃ Fx)’. See below for a full explanation. 11 As with the sentence letters of PL, we could expand the number available in QL by attaching subscripted numerals (‘a1’ ‘a2’, ‘a3’, …) if we needed to, though we will not need to in what follows.

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In addition to sentences about only one individual, we regularly use sentences about two (or maybe more) individuals. For example, ‘Mary loves John’. In such cases, we employ a similar procedure, but this time we pick out both individuals and give names to each, and pick out the two-placed predicate ‘…loves…’: Mary loves John m: Mary j: John L… … : …loves… Lmj In QL, as you can see, predicates always come before constants. This is quite different to English but it serves a very useful purpose. Suppose we want to say ‘Jane is between Jack and Jill’; in this sentence, we have a three-placed predicate to stand for a relation between the three individuals. Had we tried to represent this by having the predicate after the beginning of the sentence, we would have run into trouble. In QL, however, we have no problem: Jane is between Jack and Jill a: Jane b: Jack c: Jill B… … … : …is between …and … Babc 3. Quantifiers To complete the vocabulary of QL, we needed to add two symbols for the quantifiers. Consider again the sentence ‘Everyone is loved by someone’. The predicate is easy enough to isolate (‘…is loved by…’), but we do not have individuals which can be named by constants. Rather, our sentence says that every individual is loved by at least one individual. To express this, we need to introduce quantifiers. Quantifiers are used with variables to express the content we are looking for. The first quantifier we will consider is the universal quantifier. We use this to make hard generalisations (e.g., ‘all whales are mammals’, ‘every even number is the sum of two primes’). To move from a sentence like ‘Socrates is mortal’ (‘Ms’) to a sentence like ‘all men are mortal’, we replace the constant with a variable (which again is a lower-case Roman letter (‘x’, ‘y’, ‘z’, …).12 We then have a new sign, ‘∀’ which is attached to the variable at the beginning of the sentence to bind the variable: All men are mortal H… : …is a man M… : is… mortal (∀x)(Hx ⊃ Mx) This can be read as ‘For all x, if x is a man then x is mortal’, or just ‘For all x, if Hx, then Mx’. The second quanitifer is used for soft generalisations (e.g., ‘some men are mortal’, ‘at least one number is even’). We adopt a similar procedure, but this time we use a new sign, ‘∃’, which stands for the existential quanitifier: 12 The convention is to use letters from the end of the alphabet for variables, though we are under no obligation to, as the quantifiers make plain that these letters are variables not constants.

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Some men are mortal H…: …is a man M…: …is mortal (∃x)(Hx ∧ Mx) This can be read as ‘There exists x such that x is a man and x is mortal’, or ‘there is at least one x such that Hx and Mx’. Note that the existential quantifier means there is at least one of the things it quantifies over. This is a great deal sharper than the way we sometimes use terms like ‘some’ in English, but it carries just the same logical weight. 4. Scope of Quantifiers, Bound Variables, and Free Variables It is vital that every use of a quantifier in QL is such that the scope of the quantifier is clearly demarcated. The scope of the quantifier determines which formulas following an occurrence of quantifier are governed by it. In most cases, this will be determined by brackets: the scope of a quantifier covers the formula contained inside the outermost brackets beginning with the left bracket immediately after the formula. So, for example, the scope of the existential quantifier in the following formulas is shown in bold: (∃x)(Hx ∧ Mx) (∃x)(Hx ∧ Mx) ⊃ (∀x)Gx (∃y)(∀x)Lxy In the last example, we don’t have brackets to mark the scope of the quantifier. This is because ‘Lxy’ is an atomic formula, hence no brackets are required. It should be noted, however, that ‘Lxy’ is not a translation of any English declarative sentence. This is because, ‘x’ and ‘y’ here are variables. If ‘L’ is a translation of ‘…loves…’, then ‘Lxy’ is no more capable of being true or false in QL than ‘…loves…’ is in English. Variables are not names, they are just place holders for names (like ‘…’). To distinguish formulas containing variables from those containing names, we call them ‘open formulas’ (sometimes ‘open sentences’). If a wff is not an open formula, it is a closed formula. A closed formula is a formula which can be true or false. There are two ways of turning an open formula into a closed formula: (1) by replacing all the variables with constants; (2) by binding all the variables with quantifiers. To bind a variable with a quantifier is simply to attach a quantifier to it. For this reason we call the variables that are not bound by a quantifier in an open sentence ‘free variables’ and once we have attached a quantifier to them, we call them ‘bound variables’. If we bear this in mind, then the scope of a quantifier is easily summed up by the following rules:

1. If a quantifier is followed by an unbracketed open sentence, then the scope of the quantifier is this open sentence.

2. If a quantifier is followed by a left bracket, the scope of the quantifier extends from that bracket to the right bracket that closes it.

3. If a quantifier is followed by a tilde ‘~’, then the scope of the quantifier includes the tilde and whatever formula that tilde is the main operator of.

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5. Multiple Generality Consider the sentence ‘Everything is loved by something’.13 We now have the tools to express this formally but, to do this, we obviously need to use the two quantifiers together. It is important at this point to think carefully about how to use the quantifiers in such cases as the order of the quantifier makes a big difference to the meaning. Let us start by translating ‘…loves…’ as ‘L… …’. Clearly we need both quantifiers to express our sentence. Furthermore, the most charitable interpretation of the sentence is that it states that everything is loved by something other than itself. Therefore we will use different variables for each quantifier to make this clear. Suppose we tried: (∃x)(∀y)Lxy (read: ‘there is an x such that, for all y, Lxy’) This won’t do: what we have said here is that there is something that loves everything. But this is not the same as saying that everything is loved by something. The ‘something’ in the first case is one individual that loves everything. What we want is for everything to be loved by something (where the “somethings” in question are not all necessarily the same thing). The problem arises here because we have not paid enough attention to the logical structure of the sentence. To get it right, we need to say that everything has something that loves it: (∀y)(∃x)Lxy (read: ‘For all y, there is an x such that Lxy’) In other words, we have to make sure the right quantifier has the larger scope. Normally, as in this case, this will be reflected by the order of the quantifier expressions in the English sentence, but you have to be on your guard as this is not always the case. If I say that, in my local store, there is something for everyone, I am not saying there is one particular thing such that everyone would like that thing. Rather, I am saying that everyone is such that they will find something to their liking in the store. There are no hard and fast rules here, as the bottom line is that English allows us to express things in much more varied ways than QL. So long as you think carefully about the actual meanings of the sentences you are translating, you should be able to avoid mistakes. Tutorial Exercises (taken from Simpson, §§ 4.4 & 4.9) Using the given notation, translate each of the following into symbols. A: c: Carl p: peter f: Frank A…1 …2 : …1 annoyed …2 13 Of course, ‘everything is loved by something’ is a bit of an unusual sentence compared to our earlier ‘everyone is loved by someone’. The latter example is more complicated, however, as it restricts individuals to people. We’ll discuss this next week when we talk about domains of quantifiers but, for the moment, it is easier to leave the domain unrestricted. Notice that QL cannot distinguish active and passive forms of English sentences, so ‘a loves b’ and ‘b is loved by a’ have the same translation into QL. More advanced formal languages are able to distinguish the two, but in QL we must either treat ‘loves’ and ‘is loved by’ as distinct predicates, thus losing an explanation of the validity of arguments like ‘a loves b; a is loved by b; so a and b love each other’, or stipulate that the passive and active forms have one and the same translation (we will adopt the latter approach).

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D…1 …2 …3 : …1 discussed …2 with …3

1. Peter discussed Carl with Frank. 2. Peter was discussed by Frank with Carl. 3. Carl discussed himself with Frank but not with Peter. 4. If Frank discussed Peter with Carl and Carl discussed Frank with Peter, then Frank discussed

Carl with Peter. 5. Peter discussed himself with neither Frank nor Carl and annoyed both of them. 6. Assuming that Peter discussed Frank with Carl, then neither Frank nor Carl was annoyed by

Peter. B: e: the Empire State Building t: the Taj Mahal O…1 ….2 : …1 is older than …2

1. Something is older than something. 2. Everything is older than something. 3. Nothing is older than everything. 4. Nothing is older than itself. 5. Nothing is older than the Taj Mahal. 6. Everything which is older than the Taj Mahal is older than the Empire State Building. 7. Anything which is older than everything is older than itself.

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Interlude: Logic Test (PL Revision) This test is entirely optional, but it is recommended that you attempt it over the Easter break to get some practice of using the techniques you have learnt so far. The test will not be assessed and will not count towards your mark for this unit. The answers to the test are at the end of the book. 1. PROVIDE BRACKETS WHERE NECESSARY TO ENSURE THAT THE FOLLOWING ARE STRICTLY SPEAKING WFFS OF PL (NOTE THAT THERE IS OFTEN MORE THAN ONE WAY OF DOING THIS):

(a) ~ A ⊃ B (b) A ⊃ B ∧ ~ C ≡ A (c) A ∨ B ∨ C ∨ D ∨ E ∨ F (d) ~ ~ ~ ~ ~ ~ ~ ~ A

2. TRANSLATE THE FOLLOWING ENGLISH SENTENCES INTO PL, SHOWING THE TRANSLATIONS OF EACH ATOMIC SENTENCE:

(a) John is hungry and tired. (b) If Mary doesn’t stop talking Jane will cry. (c) Ignorance will whither unless knowledge is banished. (d) If time or money is wasted then employment costs will rise.

3. USE TRUTH-TABLES TO SHOW WHETHER THE FOLLOWING ARE TRUTH-FUNCTIONALLY TRUE, TRUTH-FUNCTIONALLY FALSE, OR TRUTH-FUNCTIONALLY INDETERMINATE.

(a) (A ≡ B) ≡ ((A ⊃ B) ∧ (B ⊃ A)) (b) (A ∨ C) ⊃ ((A ∨ B) ∨ C) (c) (A ∧ C) ⊃ ~ (B ∨ A) (d) (A ≡ G) ∧ (B ⊃ (C ∨ D))

4. USE A TRUTH-TABLE TO SHOW THAT THE FOLLOWING SEQUENT IS TRUTH-FUNCTIONALLY VALID: A ≡ B, ~ (A ≡ B) ∨ (C ∨ D) |= C ∨ D 5. CONSTRUCT PROOFS OF THE FOLLOWING SEQUENTS:

(a) A ≡ B, ~ (A ≡ B) ∨ (C ∨ D) |- C ∨ D (b) A ⊃ B, C ⊃ ~ B |- A ⊃ ~ C (c) (H ∨ I) ⊃ (J ∧ (K ∧ L)), I |- J ∧ K (d) ~ P ⊃ ~ Q, P ⊃ R, ~ (R ∨ S) |- ~ Q ∨ S

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Logic Lecture 7

Identity and Quantifier Restriction in QL

1. Identity So far, the relations we have introduced into QL have all been expressed by formulas like ‘Rab’. The relation comes before the things it relates in the same way that a predicate comes before its subject in ‘Fa’. The relation of identity, however, is treated differently (this reflects the difference in logical structure between identity and predication). To say that two things, a and b, are identical (i.e., to say that they are, in fact, the same thing), we write ‘a = b’. The negation of ‘a = b’ is written ‘~ a = b’ (there is no need to put brackets around the formula before negating it as in ‘~(a = b)’ as there can be no confusion here: ‘~ a = b’ cannot mean that the negation of ‘a’ is identical with ‘b’ as ‘a’ stands for a thing not a sentence, so ‘~ a’ is not a wff of QL). The identity sign is important for the simple reason that one person can have more than one name. Hence, we cannot assume that two constants name two distinct individuals. For example, if Clark Kent went shopping and so did Superman it does not mean that more than one person went shopping. To say that two individuals are distinct, we have to use the identity sign. The two following cases illustrate this point:

(∃x)(∃y) (Axy ∧ x = y) (∃x)(∃y) (Axy ∧ ~ x = y).

Identity becomes supremely important when we want to say that one and only one thing has such and such a property. For example, suppose we want to say that there is a present Queen of England. Obviously, we mean that there is only one person who is presently Queen of England. However, just using the existential quantifier isn’t sufficient to do this.

(∃x)Qx (Q… : …is presently Queen of England) This only tells us that at least one thing is presently Queen of England. So if there are fifteen Queens of England, our sentence will still be true. To limit our claim in the way we want, we need to write:

(∃x)(Qx ∧ (∀y)(Qy ⊃ x = y)) This states that there is at least one x which is presently Queen of England, and if anything else is presently Queen of England then that thing is identical with x. In short: There is one and only one thing which is presently Queen of England.14 2. Domains So far, we have been taking the possible values of our variables (the individuals) to simply be things. Individuals might be anything: people, dogs, plants, numbers, galaxies, or even events, spaces, points. Whatever can be the logical subject of a sentence can be treated as an individual from the point of view of logic. However, this 14 This analysis was first offered by Bertrand Russell as part of his famous ‘theory of descriptions’ first presented in his 1905 paper ‘On Denoting’ (reprinted in Russell, B. Logic and Knowledge, London: Allen and Unwin, 1956).

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leads to complications. Let us suppose that I want to translate the sentence ‘anyone bitten by a funnel-web spider will require medical treatment’. I might start with the following attempt: B… … : …is bitten by … R… : …will require medical attention F… : …is a funnel-web spider

(∀x)((∃y)(Fy ∧ Bxy) ⊃ Rx) This, however, will not do. This states that anything bitten by a funnel-web spider requires medical attention. But this is certainly not true. Only certain animals find funnel-web spider venom harmful. Cats, for example, are no more bothered by it than they would be by a bee sting. And clearly, if a confused funnel-web spider mistakenly bit a chair or leaf, no-one would think its “victims” required medical attention. To ensure that our translation is correct, then, we need to specify that we are talking about people who are bitten:

B… … : …is bitten by … R… : …will require medical attention F… : …is a funnel-web spider P… : …is a person

(∀x)((Px ∧ (∃y)(Fy ∧ Bxy)) ⊃ Rx) In cases like this, we have no choice but to introduce predicates like ‘P…’ for ‘…is a person’. In cases where all of the variables can be taken as ranging over the same things, however, we can save time and complication by restricting the domain of the quantifiers. Suppose we want to say that any funnel-web spider bitten by a funnel-web spider will suffer. We could have

B… … : …is bitten by … S… : …will suffer F… : …is a funnel-web spider

(∀x)(∀y)(((Fx ∧ Fy) ∧ Bxy) ⊃ Sx) Alternatively, however, we can simplify things greatly by assuming that we are only including funnel-web spiders among the individuals we are talking about. We do this by restricting the range of the quantifiers to the set of all funnel-web spiders. This set, is called the domain, and we write it as follows:15

{x: x is a funnel-web spider} (read: ‘the set of xs such that x is a funnel-web spider)

By specifying a domain, our translation becomes simple: Domain: {x: x is a funnel-web spider}

B… … : …is bitten by … S… : …will suffer (∀x)(∀y)(Bxy ⊃ Sx)

There are, in fact, two ways of specifying a domain. One way is to present the set by listing all of its members. For example, the two following sets are the same: {Bertrand Russell, Alfred North Whitehead} 15 A domain is also called a universe of discourse in some texts.

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{x: x is an author of Principia Mathematica} Obviously, however, we cannot list all of the members of the set of funnel-web spiders so, in such cases, we specify the set in the second way. Tutorial Exercises (taken from Simpson, §4.14, part 2) Translate each of the following into symbols. Provide the translations of each symbol. For each example, use the set of people as the domain.

1. No one who praises everyone is honest. 2. Francis Bacon is not the same person as Roger Bacon. 3. More than one person has walked on the moon. 4. Only one person has walked on the moon. 5. If Mad Max has blue eyes and Billy the Kid does not have blue eyes, then

Mad Max is not the same person as Billy the Kid. 6. Sandra is intelligent; in fact, she is the most intelligent person in the class. 7. Some retired soldiers are richer than some soldiers who have not retired. 8. Jill has only two good friends. 9. Only the good die young. 10. Jane does not love Harry but all the other girls do.

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Logic Lecture 8

Proofs in QL (I)

Just as we were able to prove the validity of arguments in PL, we can also extend the proof technique we have learnt to apply to QL. We will not go into as much detail on proofs in QL as we did with PL, but it is worthwhile finishing off the course with a brief look at how to prove things using quantifiers. A proof in QL is very similar to a proof in PL.16 We use the same introduction and elimination rules for the logical constants and we keep our rules of assumption and reiteration. All we need to add are introduction and elimination rules for the quantifiers and rules to govern use of the identity sign. We will look first at how to eliminate the universal quantifier (∀E) and then move on to existential quantifier introduction (∃I). Next week, we’ll wrap things up with the remaining rules. The rule for universal quantifier elimination is perfectly simple. If you know that everything occupies space, then you know that John occupies space. If you know that if anything is tall then it is not small, then you know that if Joe Bloggs is tall then Joe Bloggs is not small. The rule simply expresses this intuition.

∀E ONE CAN WRITE THE FORMULA ‘Φτ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT BOTH:

1. A UNIVERSALLY QUANTIFIED FORMULA ‘(∀χ)Φχ’ IS ALREADY ON THE SAME SCOPE LINE.

2. ‘Φτ’ IS THE RESULT OF REMOVING THE UNIVERSAL QUANTIFIER ‘(∀χ)’ FROM ‘(∀χ)Φχ’ AND THE REPLACEMENT OF EVERY OCCURRENCE OF THE VARIABLE ‘χ’ IN ‘(∀χ)Φχ’ BY THE SAME TERM ‘τ’.

The justification for the new line consists of an appeal to the number of the line where ‘(∀χ)Φχ’ appears and the rule, ‘∀E’.

The most important thing to remember about the rule is that it can only be applied to one universal quantifier in a formula at a time and that quantifier MUST be the main operator of the formula. Here are some examples: 16 There is, however, an extremely important difference, though we will not be concerned with it on this course. In PL, it is always possible to find out if an argument is valid by checking it on a truth-table. In QL, however, this kind of procedure cannot be done (as was famously proved independently by Alonzo Church and Alan Turing). This is because QL is not decidable, whereas PL is; this means that there is no mechanical way (like a truth-table) of testing validity in QL.

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1 (∀x)(∀y)(Bxy ⊃ Sx) P 2 (∀y)(Bay ⊃ Sa) 1, ∀E 3 Baf ⊃ Sa 2, ∀E There is no reason why we can’t replace the variable with the same name in the case of a universal quantifier (because if the statement is true of anyone then we can choose any name whatsoever). So we could have had: 1 (∀x)(∀y)(Bxy ⊃ Sx) P 2 (∀y)(Bay ⊃ Sa) 1, ∀E 3 Baa ⊃ Sa 2, ∀E The rule of existential quantifier introduction is also quite intuitive. If you know that John plays football, then you also know that someone plays football (or at least one person plays football).

∃I

One can write an existentially quantified formula ‘Φ’ on any given scope line, provided that all of the following conditions are met:

1. A FORMULA WHICH CONTAINS ONE OR MORE OCCURENCES OF THE TERM ‘τ’ APPEARS ON THAT SCOPE LINE.

2. THE VARIABLE ‘χ’ DOES NOT APPEAR IN THIS ORIGINAL FORMULA. 3. THE EXISTENTIALLY QUANTIFIED FORMULA ‘Φ’ IS FORMED BY

REPLACING ONE OR MORE OCCURENCES OF THE TERM ‘τ’ IN THE ORIGINAL FORMULA BY THE VARIABLE ‘χ’ AND WRITING ‘(∃χ)’ IN FRONT OF THE RESULT OF THE REPLACEMENT.

The justification for the new line consists of an appeal to the number of the line where the original formula containing ‘τ’ appears and the rule ‘∃I’.

Some examples make the rule clear: 1 Fa P 2 Ga P 3 Fa ∧ Ga 1,2, ∧I 4 (∃X)(FX ∧ Gx) 3, ∃I Note that we do not have to turn all the constants into variables here, we could have written: 1 Fa P 2 Ga P 3 Fa ∧ Ga 1,2, ∧I 4 (∃X)(FX ∧ Ga) 3, ∃I

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The following is also quite valid: 1 Fa P 2 Ga P 3 (∃x)Fx 1, ∃I 4 (∃y)Gy 2, ∃I 5 (∃x)Fx ∧ (∃y)Gy 3,4, ∧I Tutorial Exercises (mostly from Simpson § 5.10)

Prove the following sequents:

1. (∃x)(∃y)(∃z)Rxyz ⊃ Raas, Raaa |- (∃x) Rxxs

2. (∃x)Mx ⊃ Mk |- (∃x)Mx ≡ Mk

3. Fa, Gb |- (∃x)Fx ∧ (∃y)Gy

4. (∀x)(∀y)(Rxy ⊃ ~Ryx), Rab |- ~Rba

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Logic Lecture 9

Proofs in QL (II)

1. ∃E To understand the rule of existential quantifier elimination, we need to introduce the idea of a parameter. A parameter is sometimes called a ‘dummy’ or ‘dummy name’. Its function is more or less the same as the use of dummy names in English, like ‘Joe Bloggs’. Suppose that we know that someone (but we don’t know exactly who) has bought a season ticket for Wycombe Wanderers. Now suppose that if anyone buys a season ticket for Wycombe Wanderers then they will witness their favourite team perform very badly. As a result they will be disappointed. Then we know that someone will be disappointed. One way of presenting an argument to support this claim would be to use a dummy name like ‘Joe Bloggs’. We suppose that Joe Bloggs has bought a season ticket and then, noting that Joe will witness his favourite team perform very badly, we infer that he will be disappointed. From which we conclude that someone will be disappointed. In other words, we introduce the dummy name simply as an example of a Wycombe Wanderers fan in order to present the argument. The fact that Joe Bloggs does not actually exist does not matter as he is really just a dummy individual who we have introduced to make our point. Equally, if Joe Bloggs did exist and, as it happens, did not buy a season ticket this wouldn’t affect the validity of our reasoning as all we are doing is saying ‘suppose Joe Bloggs is the somebody in question who bought a season ticket for Wycombe Wanderers’ and then seeing what follows from this assumption. Of course, we can’t conclude anything about Joe Bloggs once we drop the assumption that he is the someone in question, but anything else that follows from Joe Bloggs buying a Wycombe Wanderers season ticket will surely follow from someone buying one. Once we have shown this, we can drop the assumption. This is just how the rule of existential quantifier elimination works: we assume that the something in question is a particular (dummy) individual and then see what follows from this assumption. We do this by introducing the dummy name (the parameter) on a scope line in the way we are now familiar with. We use the letters ‘u’, ‘v’, ‘w’, as parameters (with subscripts ‘v1’, ‘v2’, etc., if we need more) and place them to the left of the scope line marking their introduction as in the following proof: 1 (∀x)(Ax ⊃ Ba) P 2 (∃x)(Ax) P 3 u Au A 4 Au ⊃ Ba 1, ∀E 5 Ba 3,4, ⊃E 6 Ba 2,3-5, ∃E The formal statement of the rule is:

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∃E ONE CAN WRITE THE FORMULA ‘ψ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT ALL OF THE FOLLOWING CONDITIONS ARE MET:

1. AN EXISTENTIALLY QUANTIFIED FORMULA ‘(∃χ)(φχ)’ IS ALREADY ON THE SAME SCOPE LINE.

2. IMMEDIATELY TO THE RIGHT OF THE GIVEN SCOPE LINE THERE IS A SUBPROOF OF ‘ψ’ FROM ‘φτ’, WHERE ‘τ’ IS ANY PARAMETER WHICH DOES NOT OCCUR IN ‘(∃χ)(φχ)’ AND ‘φτ’ IS THE RESULT OF REMOVING THE EXISTENTIAL QUANTIFIER FROM ‘(∃χ)(φχ)’ AND REPLACING EVERY OCCURRENCE OF ‘χ’ BY ‘τ’.

3. THE SCOPE LINE OF THE SUBPROOF IS FLAGGED WITH THE SAME PARAMETER ‘τ’, AND NO FORMULA CONTAINING THE PARAMETER ‘τ’ IS MOVED ACROSS THIS SCOPE LINE.

4. ‘ψ’ DOES NOT CONTAIN ANY OCCURRENCES OF THE PARAMETER ‘τ’. THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE ON WHICH ‘(∃χ)(φχ)’ APPEARS, THE RANGE OF NUMBERS OF THE SUBDERIVATION OF ‘ψ’ FROM ‘φτ’ AND THE RULE ‘∃E’.

The rule as stated sounds pretty complicated but really it is just the form of reasoning that we have described. The important thing to remember is that the parameter marks the extent of the assumption and ensures that all reasoning using this parameter is kept within the confines of the assumption. Hence the subproof must be immediately to the right of the existentially quantified formula and we can only drop the assumption and return to our original scope line if the conclusion we have reached does not contain our dummy name (the dummy name cannot cross the scope line). For this reason, we will refer to scope lines flagged with a parameter ‘τ’ as ‘τ-barriers’. So long as these rules are followed properly, ∃E can be used several times in a proof if necessary and parameters can be introduced on secondary scope lines to the right of other parameters, as in the following proof: 1 (∃x)(∃y)Axy P 2 u (∃y)Auy A 3 v Auv A 4 (∃y)Ayv 3, ∃I 5 (∃x)(∃y)Ayx 4, ∃I 6 (∃x)(∃y)Ayx 2,3-5, ∃E 7 (∃x)(∃y)Ayx 1,2-6, ∃E

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2. ∀I The rule of universal quantifier introduction also uses parameters. They operate in a slightly more complicated way in this case, however. A few informal and formal examples will show why. The basis of the rule of universal quantifier introduction is as follows. A mathematician who wants to prove that, for example, there is no greatest prime number, will begin by saying ‘let n be any arbitrary prime number’ (‘n’ here is a dummy name) and then setting out to prove that there is always a prime number greater than n. Having proved that there is a prime number greater than any prime number n, the mathematician is justified in concluding that there is no greatest prime number (i.e. every prime is less than another prime). What is true of any completely arbitrary x is true of every x. Let us see how this technique can be used in more familiar cases. Take the following argument: Anyone who supports Wycombe Wanderers has seen their team relegated. Anyone who has seen their team relegated has experienced disappointment. ___________ Anyone who supports Wycombe Wanderers has experienced disappointment. To demonstrate the argument, we introduce a completely arbitrary individual, Joe Bloggs. If Joe Bloggs supports Wycombe Wanderers, then he has seen his team relegated. If he has seen his team relegated, he has experienced disappointment. Therefore, if he supports Wycombe Wanderers, he has experienced disappointment. As Joe Bloggs was taken as a completely arbitrary individual, we can conclude that all Wycombe Wanderers fans have experienced disappointment. The following is a formal proof of this argument: 1 (∀x)(Wx ⊃ Rx) P 2 (∀x)(Rx ⊃ Dx) P 3 u (∀x)(Wx ⊃ Rx) 1, R 4 (∀x)(Rx ⊃ Dx) 2, R 5 Wu ⊃ Ru 3, ∀E 6 Ru ⊃ Du 4, ∀E 7 Wu A 8 Wu ⊃ Ru 5, R 9 Ru 7,8, ⊃E 10 Ru ⊃ Du 6, R 11 Du 9,10, ⊃E 12 Wu ⊃ Du 7-11, ⊃I 13 (∀x)(Wx ⊃ Dx) 12, ∀I The scope line introduced at line 3 amounts to the introduction of an arbitrary individual u. ALL REASONING ABOUT u IS RESTRICTED TO THIS SCOPE LINE (and any secondary scope lines subordinate to it). Once we have completed the scope line, we move back to the left and attach the universal quantifier, replacing all occurrences of ‘u’ with the variable bound by the introduced quantifier. The reason

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why ‘u’ must not pass over the scope line is obvious when one considers the following slightly different (and invalid) argument: (P1) Graham is a Wycombe Wanderers fan. (P2) Anyone who supports Wycombe Wanderers has seen their team relegated. (P3) Anyone who has seen their team relegated has experienced disappointment. ___________ (C1) Graham has experienced disappointment ___________ (C2) Everyone has experienced disappointment. Clearly the inference from C1 to C2 is not justified. The problem was that we already knew some information about our individual, hence, we did not pick a completely arbitrary individual. The insistence on keeping the arbitrary name on a scope line of its own is needed to block this kind of invalid step. The following (mistaken) proof is a formal version of the above invalid argument: 1 Wu 2 (∀x)(Wx ⊃ Rx) P 3 (∀x)(Rx ⊃ Dx) P 4 u (∀x)(Wx ⊃ Rx) 2, R 5 (∀x)(Rx ⊃ Dx) 3, R 6 Wu ⊃ Ru 4, ∀E 7 Ru ⊃ Du 5, ∀E 8 Wu 1, R*** 9 Ru 6,8, ⊃E 10 Du 7,9, ⊃E 11 (∀x)Dx 10, ∀I The mistake here was to introduce a parameter ‘u’ which already features in a premise of the argument. As a consequence, the reiteration at line 8 (marked ***) is illicit and the subsequent reasoning is invalid. So long as one remembers that a parameter must not feature in premises or undischarged assumptions in a proof, the rule of universal quantifier introduction is fairly easy to use. The proper statement of the rule is:

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∀I ONE CAN WRITE THE FORMULA ‘(∀χ)(φχ)’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT ALL OF THE FOLLOWING CONDITIONS ARE MET:

1. IMMEDIATELY TO THE RIGHT OF THE GIVEN SCOPE LINE THERE IS A ‘τ’-BARRIER WHICH WAS NOT INTRODUCED AS THE SCOPE LINE OF AN ASSUMPTION.

2. ‘φτ’ IS A FORMULA CONTAINING ‘τ’ WHICH APPEARS AGAINST THE ‘τ’-BARRIER.

3. EVERY OCCURRENCE OF ‘τ’ IN ‘φτ’ IS REPLACED BY ‘χ’, WHERE ‘χ’ IS ANY VARIABLE WHICH DOES NOT OCCUR IN ‘φτ’.

4. THE SAME VARIABLE ‘χ’ FOLLOWS THE ‘∀’ IN ‘(∀χ)(φχ)’. THE JUSTIFICATION FOR THE NEW LINE IS THE NUMBER OF THE LINE ON WHICH ‘φτ’ APPEARS AND THE RULE, ∀I.

3. Identity The introduction and elimination rules for the identity sign ‘=’ are very, very simple indeed, so we will move straight on to stating them:

=I AT ANY POINT IN A PROOF, ONE CAN WRITE ‘τ = τ’, WHERE ‘τ’ IS ANY CONSTANT OR PARAMETER. THE JUSTIFICATION FOR THE NEW LINE IS SIMPLY THE RULE ‘=I’.

=E ONE CAN WRITE THE FORMULA ‘φτ’ ON ANY GIVEN SCOPE LINE, PROVIDED THAT ALL OF THE FOLLOWING ARE MET:

1. ON THAT SCOPE LINE THERE IS A FORMULA ‘φγ’. 2. ON THAT SCOPE LINE THERE IS A FORMULA ‘γ = τ’ OR ‘τ = γ’. 3. ‘φτ’ IS THE RESULT OF REPLACING ALL OR SOME OF THE

OCCURENCES OF ‘γ’ BY ‘τ’ IN ‘φγ’. THE JUSTIFICATION FOR THE NEW LINE CONSISTS OF AN APPEAL TO THE NUMBER OF THE LINE WHERE ‘γ = τ’ OR ‘τ = γ’ OCCURS, THE NUMBER OF THE LINE WHERE ‘φγ’ OCCURS, AND THE RULE ‘=E’.

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We will finish up with a couple of other examples of proofs that show some of the more sophisticated ways that u-barriers may be employed. 1 (∀x)(Fx ⊃ Gx) P 2 (∀x)~Gx P 3 (∀x)Fx A 4 u (∀x)Fx 3,R 5 (∀x)(Fx ⊃ Gx) 1,R 6 Fu 4, ∀E 7 Fu ⊃ Gu 5, ∀E 8 Gu 6,7, ⊃E 9 (∀x)Gx 8, ∀I 10 (∀x)Gx A 11 Ga 10, ∀E 12 (∀x)~Gx 2,R 13 ~Ga 12, ∀E 14 ~(∀x)Gx 10,11,13.~I 15 ~(∀x)Fx 3,9,14,~I 1 (∀x)(Kx ⊃ Lx) P 2 (∀x)~Lx P 3 u Ku A 4 (∀x)(Kx ⊃ Lx) 1,R 5 (∀x)~Lx 2,R 6 Ku ⊃Lu 4, ∀E 7 ~Lu 5, ∀E 8 Lu 3,6, ⊃E 9 ~Ku 3,7,8,~I 10 (∀x)~Kx 9, ∀I The technique emplyed in the last proof is especially useful when seeking to introduce a quantifier that has wide scope over a negation operator. To test your skills with quantifiers to the full, you might like to try and prove the following quantifier equivalences:

1. (∃x)Fx ≡ ~(∀x)~Fx 2. ~(∃x)Fx ≡ (∀x)~Fx 3. ~(∀x)Fx ≡ (∃x)~Fx 4. ~(∃x)~Fx ≡ (∀x)Fx

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Tutorial exercises: proofs in QL Prove the following sequents of QL:

5. (∀x)(∀y) Rxy, (∀x) Rxx ⊃ ~Fc |- ~Fc 6. (∀x) (Jx ⊃ Lx), (∀w) (~Qw ≡ Lw) |- ~(Jb ∧ Qb) 7. (∃x) Px ⊃ Sc, (∃x) (Txx ∧ (∃y) (Py ∧ ~Jy)) |- Sc 8. (∃y) (My ∧ Ry), (∀y) (Lyn ⊃ ~ (Ry ∨ Cy)) |- (∃w) ~ Lwn

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Appendix A

Revision of proofs in PL_ All of the following are taken from Simpson §§ 3.10, 3.19, & 3.23 Construct proofs of the following sequents (note: some require assumptions, some do not require assumptions, and some require categorical proofs – you should think carefully about what kind of proof is likely to be called for before you put pen to paper):

1. A, B, A ⊃ ~ ~ (B ⊃ ~ ~ C) |- C 2. P, P ⊃ ~ ~ ~ S, A ≡ (R ∨ S), (A ∨ B) ≡ (P ∧ Q), ~ S ⊃ F, F ⊃ R |- (P ∧ Q) ∧ R 3. ~ (A ∨ B) |- ~ A 4. N ⊃ O, O ≡ P, N ≡ S |- S ⊃ P 5. A ≡ B, A ∨ B |- A ∧ B 6. M |- ~ M ⊃ O 7. |- ~ F ⊃ (F ⊃ G) 8. |- ~ ((F ∧ G) ∧ (~ F ∨ ~ G))

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Appendix B PL Revision, Logic Test Answers 1. (n.b there are several ways of answering these qs.)

(e) (~ A ⊃ B) (f) (A ⊃ ((B ∧ ~ C) ≡ A)) (g) (((A ∨ B) ∨ (C ∨ D)) ∨ (E ∨ F)) (h) ~ ~ ~ ~ ~ ~ ~ ~ A

2. (a) John is hungry and tired. H: John is hungry T: John is tired J ∧ H (b) If Mary doesn’t stop talking Jane will cry. M: Mary stops talking J: Jane will cry ~ M ⊃ J (c) Ignorance will wither unless knowledge is banished. I: Ignorance will wither K: Knowledge is banished I ∨ K or you might prefer: ~ K ⊃ I (d) If time or money is wasted then employment costs will rise. T: Time is wastedM: Money is wasted E: Employment costs will rise (T ∨ M) ⊃ E 3.

(a)

A B (A ≡ B) ≡ ((A ⊃ B) ∧ (B ⊃ A)) T T T T T T T T F F T F F T F T F T T F F F F T T T T T 2 1 4 3 5

truth-functionally true. (b)

A B C (A ∨ C) ⊃ ((A ∨ B) ∨ C) T T T T T T T T T F T T T T T F T T T T T T F F T T T T F T T T T T T F T F F T T T F F T T T F T F F F F T F F 2 1 4 3 truth-functionally true

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(c)

A B C (A ∧ C) ⊃ ~ (B ∨ A) T T T T F F T T T F F T F T T F T T F F T T F F F T F T F T T F T F T F T F F T F T F F T F T T F F F F F T T F 2 1 3 4 truth-functionally indeterminate A B C D G (A ≡ G) ∧ (B ⊃ (C ∨ D)) T T T T T T T T T T T T T F F F T T T T T F T T T T T T T T F F F F T T T T F T T T T T T T T F T F F F T T T T F F T T F F F T T F F F F F F F T F T T T T T T T T F T T F F F T T T F T F T T T T T T F T F F F F T T T F F T T T T T T T F F T F F F T T T F F F T T T T F T F F F F F F T F F T T T T F F T T F T T T F T T T T F T T F T F F T T F T T F F T T T T F T F T T F F T T F T F T F T T T T F T F F T F F F F F T F F F T F F F F F T T T F F T T F F T T F T T T T F F T F T F F T T F F T F F T T T T F F F T T F F T T F F F T F T T T T F F F F T F F T F F F F F F T T T F 4 1 2 3 truth-functionally indeterminate

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4. A B C D A ≡ B, ~ (A ≡ B) ∨ (C ∨ D) |= (C ∨ D) T T T T T F T T T T T T T F T F T T T T T T F T T F T T T T T T F F T F T F F F T F T T F T F T T T T F T F F T F T T T T F F T F T F T T T T F F F F T F T F F F T T T F T F T T T F T T F F T F T T T F T F T F T F T T T F T F F F T F T F F F F T T T F T T T T F F T F T F T T T T F F F T T F T T T T F F F F T F T F F F

1 2 4 1 3 1 truth-functionally valid 5.

(e) A ≡ B, ~ (A ≡ B) ∨ (C ∨ D) |- (C ∨ D) 1 A ≡ B P 2 ~ (A ≡ B) ∨ (C ∨ D) P 3 C ∨ D A 4 ~ (A ≡ B) A 5 ~ (C ∨ D) A 6 A ≡ B 1, R 7 ~ (A ≡ B) 4, R 8 ~ ~ (C ∨ D) 5,6,7, ~I 9 C ∨ D 8, ~E 10 C ∨ D 2,3-3,4-9, ∨E

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(f) A ⊃ B, C ⊃ ~ B |- A ⊃ ~ C

1 A ⊃ B P 2 C ⊃ ~ B P 3 A A 4 C A 5 A 3, R 6 A ⊃ B 1, R 7 B 5,6, ⊃E 8 C ⊃ ~ B 2, R 9 ~ B 4,8, ⊃E 10 ~ C 4,7,9, ~I 11 A ⊃ ~ C 3-10, ⊃I

(g) (H ∨ I) ⊃ (J ∧ (K ∧ L)), I |- J ∧ K 1 (H ∨ I) ⊃ (J ∧ (K ∧ L)) P 2 I P 3 H ∨ I 2, ∨I 4 J ∧ (K ∧ L) 1,3, ⊃E 5 J 4, ∧E 6 K ∧ L 4, ∧E 7 K 6, ∧E 8 J ∧ K 5,7, ∧I

(h) ~ P ⊃ ~ Q, P ⊃ R, ~ (R ∨ S) |- ~ Q ∨ S 1 ~ P ⊃ ~ Q P 2 P ⊃ R P 3 ~ (R ∨ S) P 4 Q A 5 ~P A 6 ~ P ⊃ ~ Q 1, R 7 ~ Q 5,6, ⊃E 8 Q 4, R 9 ~ ~ P 5,7,8, ~I 10 P 9, ~E 11 P ⊃ R 2, R 12 R 10,11, ⊃E 13 R ∨ S 12, ∨I 14 ~ (R ∨ S) 3, R 15 ~ Q 4,13,14, ~I 16 ~ Q ∨ S 15, ∨I [brain teaser: (h) can be proved in 11 lines – can you find a shorter proof than the one given here?]