Department of Philosophy, WITS PHIL 3009 SYMBOLIC … logic... · Department of Philosophy, WITS PHIL 3009 SYMBOLIC LOGIC COURSE PACK Lecturer: MURALI RAMACHANDRAN INTRODUCTION: DEDUCTIVE

Department of Philosophy, WITS

PHIL 3009 SYMBOLIC LOGIC COURSE PACK Lecturer: MURALI RAMACHANDRAN

INTRODUCTION: DEDUCTIVE AND INDUCTIVE ARGUMENTS

An argument is collection of propositions, one of which—the conclusion—is presented as

being supported (backed-up) by the others—the premises.

Deductive validity

A (deductively) valid argument is one where it is impossible for the premises to be true

without the conclusion also being true; i.e. where it would be contradictory to affirm the

premises but deny the conclusion. When an argument is valid, we say the premises entail the

conclusion.

Any argument that is not valid is said to be invalid.

Examples:

Argument A

Premises: (1) Tracy is a vegetarian or a smoker.

(2) If she is a smoker, she lied to her dad.

(3) Tracy has never lied to her dad.

Conclusion: (4) So, Tracy is a vegetarian. Valid

Argument B

Premises: (1) Vince is a nerd if Brian is.

(2) Brian isn‟t a nerd unless he supports United.

(3) Brian doesn‟t support United.

Conclusion: (4) Hence, Vince is a nerd. Invalid

Inductive strength

An inductively strong argument is one whose premises would provide positive support for

the conclusion if they were true—the premises render the conclusion more likely.

Argument C

Premises: (1) Kev is an animal-rights activist and

Beth is a butcher.

Conclusion: (2) So, if either is a vegetarian, Kev is.

Inductively strong

Argument D

Premises: (1) Malcolm is an accountant.

(2) Beth was nearly bored to death by some

accountants at a party once.

Conclusion: (3) So, Beth will (probably) find Malcolm

boring too. Inductively weak

Dept of Philosophy, WITS SYMBOLIC LOGIC Page 2

Differences between validity and inductive strength

An argument is (deductively) valid or inductively strong NOT in virtue of the truth of its

premises or conclusion, but in virtue of how the premises are related to the conclusion, i.e.

what degree of support they would provide for the conclusion. So, e.g. an argument with false

premises may still be valid or inductively strong—consider (1) of argument A.

Whether an argument is valid or not is knowable a priori. But inductive strength depends

on background knowledge. The force of argument C, for example, stems from our knowledge

of butchers and animal-rights activists. This suggests that inductive strength is not generally

known a priori.

Adding further premises to a valid argument cannot make it invalid. In contrast, whether an

argument is inductively strong or weak is defeasible: the „strength‟ of the argument may be

altered by adding further premises. E.g. suppose one added the further premise that Beth

comes from a long line of vegetarian butchers in argument C. The conclusion does not seem as

plausible as before. Hence, inductive strength, unlike validity, admits of degrees.

Logical form

Certain valid arguments are valid purely in virtue of their „shape‟. For example, this is the

shape (form) of Argument A:

Premises: V or S

If S, then L

Not-L

Conclusion: V

Any argument of the same form is guaranteed to be valid. Thus, this is called a valid logical

form.

Formal logic is the study of logical form and it is this we shall be concerned with hereafter in

this course, since it provides a fundamental and relatively easy starting point.

Exercises

1. For each of the following arguments state whether it is deductively valid, inductively

strong, or inductively weak giving your reasons.

(a) Lucy won‟t find Logic difficult if she‟s good at Maths, and she got top marks in her

Maths test. So, she‟s not going to find Logic difficult.

(b) Raju‟s parents are strict vegetarians. Now, he wouldn‟t eat meat unless they did. So, he

too must be a vegetarian.

(c) It is highly likely that Sabrina has musical talent. For, the indications are that musical

talent is largely inherited and Sabrina‟s sisters, brothers, parents and grandparents are

all excellent musicians.

(d) Willard always leaves food on his plate when he is ill. He must have been ill last

Monday: he came in here that day (after being fired), had one mouthful of stew and

promptly left!

2. Give an example of strong inductive argument. Now add a premise to make it weak

inductive argument.


TRUTH-FUNCTIONALITY

Sentence connectives

A sentence connective operates on sentences to form compound (or complex) sentences.

E.g. „… and …‟ operates on pairs of sentences to form compound sentences of the form

„A and B‟; „If … then …‟ operates on pairs of sentences to form sentences of the form „If A

then B‟.

NOTE on negation: „It is not the case that …‟ is treated as a sentence connective even

though it just operates on single sentences. (It is called a one-place connective).

Truth-functional connectives

A sentence connective is truth-functional if the truth-value of any sentence it gives rise to

is wholly determined by the truth-values of the simpler sentences it combines. For example,

conjunction („and‟, &), disjunction („or‟, ) and negation („not‟, ) are all truth-functional

connectives. They are fully defined in the following truth-table:

A

B

A and B

A & B

A or B

A B

not-A

A

T T T T F T F F T F

F T F T T

F F F F T

Explanation

The „A‟ and „B‟ stand for propositions (or declarative sentences—sentences that can be true

or false). „T‟ and „F‟ stand for the truth-values true and false, respectively.

It is assumed that a proposition must be true or false: it can’t be neither.

Now, consider e.g. the truth table for „‟ („or‟). The first row indicates that when A and B are

both true, „A B‟ is true; the second row indicates that when A is true and B is false, „A B‟ is

true, and so on. The only circumstances in which „A B‟ is false is when both A and B are false.

A non-truth-functional connective

„A because B‟ is an example of a sentence connective that is not truth-functional, i.e. its

truth-value is not determined solely by the truth-values of A and B. It can be true when A and

B are both true, as in e.g. „The fire spread because the doors were not closed‟; but it may also

be false when A and B are both true, as in e.g. „The fire spread because Oswald shot

Kennedy‟.

Complex sentences

A simple (or atomic) sentence is one that does not have any connectives (e.g. , , &) in

it. A complex sentence is one that does.

A connective may operate on complex sentences to generate even more complex

sentences, e.g. „(A B) & C‟ is also a sentence.

The brackets (parentheses), „(‟ and „)‟ are required to avoid ambiguity. To give a simple

example, compare the English sentences “Arnie didn‟t go to the party but Belinda did” and

“Arnie and Belinda didn‟t both go to the party”. The first sentence affirms that Belinda didn‟t


go to the party, but the second would be true if Belinda went and Arnie didn‟t. The sentences

would be expressed in our logical language as: „ A & B‟ and „ (A & B)‟, respectively.

Working out the truth-value of complex sentences

Great news! We can work out the truth-value of any complex sentence if we know the truth-

values of the simple sentences it contains.

Example: What is the truth-value of „(A B) & C‟ if B and C are true and A is false?

We begin by wrting down the truth-values of the simple sentences

(and numbering the steps below—this lets me know the order in

which you have done things). We have to work out the truth-value

of the sentence in brackets first. We begin by putting the truth-

value of „ B‟ under the „‟, then we can work out the truth-value of „A B‟:

Since „A‟ is false, and „ B‟ is false, the truth-value of „A B‟ is

the truth-value of „false false‟, which according to the truth table

of „‟ is false.

The truth-value of the whole sentence „(A B) & C‟ is the value under the main connective, in

this case, the „&‟. The main connective of a sentence is the one with the fewest brackets around it.

(If there is a tie, it is the one that is not „‟.)

The truth-value under the „&‟ is determined by the truth-values of

„(A B)‟ and „C‟; i.e. we want the value of „false & true‟;

according to the truth table definition of „&‟, this is false. So, the

truth-value of the whole sentence is FALSE.

Another example. Work out the truth value of (A & B) if A and B are both true.

Again, we begin by putting the values of the simple sentences below

them (steps 1 and 2). Next comes the „‟ inside the bracket, then the

„&‟, then finally the „‟ outside the bracket, which is the main

connective.

NOTE: step 5 is determined by looking at step 4, not step 1! This is because the truth value of

the bracketed sentence „(A & B)‟ is the value under the „&‟.

Exercises

If A, B and C are true and D and E are false, what are the truth-values of the following

sentences?

SENTENCE T/F

a. C

b. (B & A)

c. (( D & C) E)

d. (A D) & (A & D)

e. (E ( A & C))

(A B) & C F T T

1 2 3

(A B) & C F F F T T

1 5 4 2 3

(A B) & C

F F F T F T

1 5 4 2 6 3

(A & B) T T F F T

5 1 4 3 2


Working backwards

In some cases, we can work „backwards‟ to find the truth-value of atomic sentences from

the truth-value of complex sentences.

Example: Suppose „ C (S P)‟ is false. What must the values of C, S and P be?

Step 1 is false. The truth table for „‟ shows that „‟ is false only

when both disjuncts are also false; hence steps 2 and 3. If „ C‟ is

false, C must be true; thus, step 4. Finally, again using the table for

„‟, we see that it is false only when both sides are false; hence

steps 5 and 6. So, the whole sentence is false only if C is true and S and P are false.

Exercises

1. If „P & (Q & R)‟ is true, and R is true, what are the truth-values of P and Q?

2. If „ (A & (B A))‟ is false, what are the values A and B?

Another logical connective: ‘’

The truth table below introduces and defines a new connective, „‟:

A B A & B A B A A B

T T T T F T

T F F T F F

F T F T T T

F F F F T T

An „arrow‟ statement, „A B‟, is FALSE only when A is true and B is false; it is TRUE

in all other circumstances. (We‟ll consider what „A B‟ means, what sorts of sentences of

ordinary language it is used to translate or represent shortly.)

Exercises

1. What are the truth values of the following sentences if P is true, and Q and R are false:

1a. P (Q R)

1b. (( Q & (Q P))

2. What are the truth values of P, Q and R if the following sentence is false:

( P Q) (P R)

Using truth tables to determine logical equivalence, tautologies, etc.

Two sentences are logically equivalent if they cannot differ in truth value: any situation

where one is true is a situation where the other is true also. We can establish whether a pair of

SL-sentences are logically equivalent by constructing truth tables. Two SL-sentences are

logically equivalent if the values under their main connectives in the truth table are exactly

the same.

A logically true sentence (a tautology) has „T‟ under its main connective in every row. A

logically false (or self-contradictory) sentence has „F‟ under its main connective in every row.

C (S P) F T F F F F

2 4 1 5 3 6


A sentence that is neither logically true nor logically false s said to be logically indeterminate

(or contingent).

We can also establish whether a set of SL-sentences is consistent, i.e. whether the

sentences could all be true simultaneously, by constructing a truth table. The set is consistent if

there is a row in the truth table which has a T under the main connective of every sentence in

the set. (It is inconsistent otherwise.)

Examples:

Which of the following pairs of sentences are logically equivalent?

(1a) E E; E (1b) K L; K & ( K L)

(1a) The first section of the table depicts the possible combination

of truth values of the atomic sentences involved. The truth values

under the main connectives match each other. So, the sentences in

(1a) are logically equivalent.

(1b) Again, the first section of the table depicts the

possible combination of truth values of the atomic

sentences. Notice the order in which they are

written: it is important to have such a system to

ensure that all the possible combinations are

covered. These sentences are NOT logically

equivalent. For example, when K is false and L is

true (see third row), the first sentence is true but the second is false.

More examples: Construct truth tables to determine whether the following sentences are

logically true, logically false, or logically indeterminate:

(2a) L (L L)

(2b) (L P) L

L P L (L L) (L P) L

T T

T F

F T

F F

Complete the above truth table and answer the question!

Exercises

1. Construct a full truth table to establish whether the following sentence is logically true,

logically false, or logically indeterminate:

(A & B) A

E E E E

T

F

T F F

F T T

F

T

K L K L K & ( K L)

T T

T F

F T

F F

T

T

T

F

4

T F T

F F F *

F T T *

F T T

3 1 2


2. Construct a full truth table to establish whether the following pair of sentences are logically

equivalent or not:

L M; M L

Sentential Logic (SL) It is time to introduce you to a simple logical language, Sentential Logic (also referred to as

propositional logic), which consists of atomic sentences and rules for combining sentences

with truth-functional sentence connectives. In such a language it is easy to evaluate the truth

values of complex sentences and to establish the validity or invalidity of arguments. [Our

ultimate goal is to evaluate ordinary-language arguments by way of representing them in

Sentential Logic.]

Vocabulary Atomic sentences: A, B, C, ..., Z, A1, ..., Z1, A2, ..., Z2, ...

Truth-functional connectives:

NAME SYMBOL INTERPRETATION/COMMENTS

NEGATION , ( A) It is not the case that A

CONJUNCTION &, (A & B) A and B (A and B are called conjuncts)

DISJUNCTION (A B) A or B (A and B are called disjuncts)

IMPLICATION , (A B) If A, then B (This is also called a material

conditional; A is called the antecedent and B

the consequent of the conditional)

EQUIVALENCE , (A B) A if and only if B (This is also called a material

bi-conditional)

Punctuation marks: Left and right brackets (parentheses): ( , )

Formation Rules

What counts as an SL-sentence?

(a) Every atomic sentence is an SL-sentence.

And for any SL-sentences A and B, (b)-(f) are also SL-sentences:

(b) A; (c) (A & B); (d) (A B); (e) (A B); (f) (A B)

Bracket convention:

The outermost brackets of a sentence may be dropped. So, e.g. „A (B & C)‟ in place of

„(A (B & C))‟ is permissible, but „ A B‟ in place of „ (A B)‟ is not permissible.

Example: Explain, using the formation rules above, why „ B ( B & A)‟ is an SL-sentence.

i. By clause (a), A and B are SL-sentences.

ii. By clause (b), B is an SL-sentence.

iii. By clause (c), ( B & A) is an SL-sentence.

iv. By clause (e), ( B ( B & A)) is an SL-sentence.

v. Finally, by the bracket convention, B ( B & A) is an SL-sentence.


Truth-table definition of the connectives

As before, the connectives of SL are fully defined by truth tables. In addition to the connectives

introduced earlier, we have the bi-conditional, „‟:

A A & B A B A B A B

F T T T T T T T T T T T T T

T F T F F T T F T F F T F F

F F T F T T F T T F F T

F F F F F F F T F F T F

Sentential Logic: Translation Before we can evaluate ordinary-language arguments, we need to know how to translate

complex English sentences into SL. Here is an important distinction to bear in mind before we

consider some of the more common translation rules.

Distinction between what is said (asserted) and what is implied or merely conveyed

Our intuitions about translation can be misleading.We must distinguish between the literal

content of an utterance and what is implied by the utterance. Consider e.g. an utterance of the

sentence “I brushed my teeth and took off my specs”. The speaker implies that she brushed

her teeth before taking off her spectacles—but this is not part of what is literally said. For, it

would not be contradictory if she added, “… but not in that order”. The implication is thus

cancelable (the phrase is H.P. Grice‟s). What is literally affirmed is not cancelable in this

sense; e.g. it would be contradictory if she had instead added “… but I didn‟t brush my teeth”.

Her original utterance, “I brushed my teeth and took off my specs”, is literally true if, and only

if, she did both those things (but not necessarily in that order). Hence, the earlier truth-table

definition of “and” stands.

Likewise, if someone says, “Blair is either very rich or very clever”, she may imply that he

is not both, but this is not part of what is literally asserted: it would not be contradictory for

her to add, “In fact, he is both.”

Another example. Suppose someone says, “If I win the lottery, I will retire”. One might

well take the speaker to be implying that if she doesn’t win the lottery, she won’t retire. But

this implication is cancelable: she could have added, “Come to think of it, I might retire even if

I don‟t!”

Cancelability is a useful simple test for distinguishing what is literally said from what is

merely implied or conveyed.

Translation of ‘but’

Consider the statement, „Annie is poor but she is happy‟. The use of the word „but‟ implies

that the two conjuncts in some tension which each other: that Annie‟s being rich is somewhat

surprising or unexpected in light of her being poor. But, the statement is true so long as both

conjuncts are.

More generally, „A but B‟ has the same truth conditions as „A and B‟, and simply gets

translated into SL as „A & B‟.


Translation of If-statements

A statement of the form „If A, then B‟ is definitely false if A is true and B is false—e.g. „If

Annie goes to the party, Bill will‟ is definitely false if it turns out that Annie goes and Bill

doesn’t. Note that „A B‟ is also false when A is true and B is false.

One of the uses of „‟ in elementary logic is to translate If-statements, e.g. statements of

the form “If A, then B” and “A if B”.

SIMPLE RULE FOR TRANSLATING IF-STATEMENTS WITH „‟:

The If-part goes in front of the arrow.

o So, „If A then B‟ gets translated as „A B‟,

o whereas „A if B‟ gets translated as „B A‟

It may seem counterintuitive to count „If A, then B‟ as true when A is false. But think of

this translation as charitable way of accommodating the assumption that every sentence must

be either true or false.

Translation of ‘Only If’-statements

Now, ask yourself, when is „Annie will go to the party only if Bill does‟ definitely false?

— Answer: if it turns out that Annie goes and Bill doesn‟t.

More generally, a statement of the form „A only if B‟ is definitely false in the same

circumstances that „If A, B‟ is definitely false: i.e. when A is true and B is false.

So, „A only if B‟ gets translated into SL as „A B‟ as well.

So, as far as SL is concerned, „If Annie goes to the party, Bill will‟ and „Annie will go to

the party only if Bill does‟ say the same thing—at any rate, have the same truth conditions.

This will seem counterintuitive: surely the first implies that Annie‟s going would be responsible

for Bill‟s going, whereas the only-if statements implies that Annie‟s going would depend on

Bill‟s going? These casual implications are lost in the SL-translations.

Example. Let us take (1) as our candidate for translation:

(1) If Adam doesn‟t pass his driving test and Bianca does, he will get upset.

The first step is to assign a (capital) letter to each of the „atomic’ propositions—the simplest,

„non-negative‟ propositions—that the sentence involves. There are three atomic propositions

in this case:

A : Adam passes his driving test.

B : Bianca passes her driving test.

U : Adam will get upset.

Using these letters, we see that what sentence (1) affirms (getting ungrammatical awhile) is:

(1a) If not-A and B, then U.

Now, (1a) contains three connectives: „if ... then‟, „not‟ and „and‟. Which one is the main

connective? I.e., is the sentence fundamentally an „if ... then‟-sentence, a „not‟-sentence or an

„and‟-sentence? Clearly, it is the first. So the main connective of the SL-translation of (1a) will

be „‟; that is, the SL-translation will be of the form: *** U.

Next, what goes in place of ‘***’? The SL-translation of „not-A and B‟, which is: „ A & B‟;

and we need brackets around this, because „&‟ is not the main connective of the whole

sentence. So the SL-translation of (1) is:

(1SL) ( A & B) U


Translation Exercises 1

Using A, B and U to stand for the same propositions as above, symbolize the following

statements as sentences of SL:

a. Bianca passed her driving test but Adam didn‟t.

b. Neither Adam nor Bianca passed their driving test.

c. Adam will not get upset if he passes.

d. Adam will get upset only if he fails and she passes.

Translation of ‘Unless’-statements

The statement „The patient will die unless the doctor arrives soon‟ basically affirms that the

patient will die if the doctor does not arrive soon. More generally, „A unless B‟ should be read

as „A if not-B‟.

Given our simple rule for translating if-statements (that the if-bit goes in front of the

arrow), the SL-translation is „A if not-B‟ is „B A‟

SIMPLE RULE FOR TRANSLATING UNLESS WITH AN ARROW: JUST REPLACE IT WITH „IF-NOT‟

o So, e.g. „Unless A, B‟ should be translated as „If not-A, B‟, i.e. „A B‟

„A B‟ is actually equivalent to „A B‟—(check it!).

So, AN EVEN SIMPLER RULE: JUST TREAT „UNLESS‟ AS DISJUNCTION, „‟.

o So, e.g. both „A unless B‟ and „Unless A, B‟ can be translated as „A B‟

„But‟, „If‟, „Only if‟, „if and only if‟ and „Unless‟ are some of the common sentence

connectives in English; you are going to have to figure out how to best translate the less

common ones yourself!

Translation Exercises 2

Symbolize the following statements as sentences of SL, using the suggested abbreviations:

KEY

A : Anna went to the party. D : Dora went to the party.

B : Barry went to the movies. E : Anna enjoyed the party.

C : Cheryl got drunk at the party.

(a) Dora didn‟t go to the party.

(b) Anna went to the party but didn‟t enjoy it.

(c) If Anna went to the party, Barry went to the movies.

(d) Either Anna or Dora went to the party.

(e) Anna and Dora didn‟t both go the party.

(f) Cheryl got drunk at the party if Anna went to it.

(g) Anna went to the party only if Barry went to the movies.

(h) Dora went to the party so long as Anna did.

(i) Cheryl didn‟t get drunk at the party unless Anna or Dora went to it.


Using Truth Tables to Establish SL-Validity We are going to see how to establish whether an argument is valid or not using a complete

truth table. Later, we will consider a simple shortcut method that can save a lot of time.

Consider the following argument:

Sarah is in Paris if Dave is. She is in Cambridge if Dave is not in Paris. As we know,

she definitely is not in Cambridge. So, both Sarah and Dave are in Paris.

(S = Sarah is in Paris; D = Dave is in Paris; C = Sarah is in Cambridge)

The full truth table method

The first step is to translate the argument into Sentential Logic:

D S, D C, C |= S & D

(„|=‟ is the entailment sign: it separates the premises from the conclusion).

A B A & B A B A A B

T T T T F T

T F F T F F

F T F T T T

F F F F T T

The figure below shows a full truth table for the argument (using the truth table definitions

of the connectives above):

C D S D S D C C S & D

T T T T F T T F T

T T F F F T T F F

T F T T T T T F F

T F F T T T T F F

F T T T F T F T T *

F T F F F T F T F

F F T T T F F T F

F F F T T F F T F

Note: The left-hand columns list the sentence letters involved in alphabetical order. The

truth values in the columns below are arranged so that all the possible combinations are

exhausted. The conclusion of the argument appears in the right-most column.

The asterisks indicate the rows in which the ALL the premises are true. (In this case, there

is only one such row, but in other cases there may be many, or even none.)

The table shows that the argument is VALID: because on all the rows in which the

premises are true, the conclusion is also true.

An argument is only INVALID if there is at least one row in its truth table where ALL the

premises are true but the conclusion is false. (Note: if there are some or many rows with true

premises and true conclusion and just one row with true premises and false conclusion, the

argument is still INVALID.)

Note: if there is NO row with all the premises coming out true, then the argument is

VALID.


Exercises on full truth table method

Translate the following arguments into sentences of Sentential Logic and then construct full

truth tables to establish whether or not they are valid:

(1) Either Ernie has a cold or he enjoys blowing his nose. If the latter, he is weird. Hence,

since Ernie does not have a cold, we can conclude that he is weird.

(C = Ernie has a cold; E = Ernie enjoys blowing his nose; W = Ernie is weird)

[Translation of ‘since’. You can translate ‘Since A, B’ as ‘A B’; but, alternatively—

and this is what I would recommend you do generally: take the since-bit, ‘A’, to be an

additional premise and take the conclusion of the argument to simply be ‘B’.]

(2) Bianca won‟t buy a car unless she passes her test. But she‟ll pass only if her partner

does. So, Bianca will buy a car if her partner passes the test.

(B = Bianca will pass; P = Bianca‟s partner will pass; C = Bianca will buy a car)

(3) Lord Dweeb is not guilty if the butler is lying. But if Dweeb is guilty, he was being

blackmailed. So, either the butler is telling the truth or Dweeb was being blackmailed.

(G = Dweeb is guilty; B = Dweeb was being blackmailed; L = the butler is lying).

The shortcut method The strategy of the shortcut method is try and make the premises true and the conclusion

false by assigning truth values. If you are thereby forced into assiging contradictory truth

values somewhere along the line, you have shown the argument to be valid, because you have

shown it is impossible for the premises to be true and the conclusion false. But if can assign

the truth values without any resulting contradiction, you have shown the argument to be

invalid.

Let us see the method used with the earlier example.

Always begin by assigning the premises true

and the conclusion false and number the

assignments underneath.

Then see if any other assignments are forced

upon you. In this case, step 3 dictates that C is

false, so we can put that down (step 5).

Steps 2 and 5 force you to assign false to

„ D‟ (step 6). This means that D must be true

(steps 7 and 8).

Steps 7 and 1 force you to assign true to S

(steps 9 and 10).

D S, D C, C |= S & D T T T F

1 2 3 4

D S, D C, C |= S & D T T F T F

1 2 5 3 4

D S, D C, C |= S & D T T F T F T F T

7 1 6 2 5 3 4 8

D S, D C, C |= S & D T T T F T F T T F T

7 1 9 6 2 5 3 10 4 8


But now we have arrived at a contradiction: for steps 10, 4 and 8 cannot all be correct

(true & true should be true, not false). So, we have shown that it is impossible for the

premises to be true and the conclusion false: i.e. the argument is VALID.

Another example: Determine whether or not the following SL-argument is valid:

R S, S E, E |= R

Here are the various stages:

First, assign the premises true and the conclusion false.

(Notice that we assign „E‟ a truth value at step 3. It has no

connective but it is still a premise, so it must be assigned true.)

Next, since step 3 assigns true to E, we can write this down

under the E in the second premise (step 5). Steps 2 and 5 can

only be correct if S is true (step 6).

Now we can put true under S in the first premise (step 7).

We have R as false at step 4, so we can put this down under R

in the first premise (step 8).

We have finished assigning the values. There is no contradiction anywhere: F T should

be true, as the first premise has it; likewise, T T should be true, as the second premise has it;

and the assignments to E and R do not conflict with any of the earlier values. So, we have

shown it is possible for the premises of this argument to be true while the conclusion is false.

That is all that is required for the argument to be INVALID.

When an argument is invalid, you need to specify truth values of the atomic

sentences which result in true premises and false conclusion.

o So in the above example, final answer should be something like this: When R

is true and E and S are false, the premises are true but the conclusion is false,

so the argument in INVALID.

The usefulness of the shortcut method is clear when we have an argument with four or

more atomic sentences. Consider e.g.: G F, B F, B R |= R G

Steps 7 and 12 contradict step 1. So, it

is not possible for the premises to be true

while the conclusion is false. Hence, the

argument is VALID.

(A full truth table for the above argument, with 5 sentence letters, would require 32 rows! The

shortcut method makes the task easy.)

Sometimes, the initial assignment of values does not fully determine the truth values of all

the atomic sentences. Consider e.g.: (P W) (P & W) |= P

This is far as one can go from the initial assignment

(steps 1 and 2). Step 5 follows from step 3 because

both F T and F F are true.

R S, S E, E |= R T T T F

1 2 3 4

R S, S E, E |= R T T T T T F

1 6 2 5 3 4

R S, S E, E |= R F T T T T T T F

8 1 7 6 2 5 3 4

G F, B F, B R ╞═ R G

F T F T T T F T F T F F

7 1 12 10 2 11 9 3 8 5 4 6

(P W) (P & W) ╞═ P

F T T F F

3 5 1 4 2


But in this case, we have enough information to see that the argument is invalid. For, step

5 guarantees the correctness of step 1—because a disjunction (an „or‟ statement) is true if

either disjunct (i.e. either side) is true. So we need go no further. We have shown that it is

possible for the premises to be true and the conclusion to be false; namely, when „P‟ is false,

regardless of the value of „W‟. Hence, the argument is INVALID.

Exercises on the shortcut method:

1. Use the shortcut method to establish whether the following SL-arguments are valid or

invalid:

(a) J M, J ( J & M) |═ M (M & J)

(b) (M & J), J K |= K M

2. Translate the following arguments into Sentential Logic, and then determine whether

they are valid or invalid using the shortcut method:

2a If Blair resigns and there is an immediate general election, then Cameron will become

Prime Minister if the newspapers support him. Clearly, if Blair resigns, there will be an

immediate general election. And, equally clearly, the newspapers will support Cameron

if Blair resigns. So, unless Blair does not resign, Cameron will become Prime Minister.

(Use the following key: B = Blair will resign; C = Cameron will become PM;

E = There will be a general election; N = The newspapers will support Cameron)

2b Belinda is not guilty if Andrew‟s testimony is true. And if Belinda is not guilty,

Charlene lied. Since Charlene did not lie, Andrew‟s testimony is not true.

(B = Belinda is guilty; A = Andrew‟s testimony is true; C = Charlene lied)

2c. Brains‟ plan will work if and only if Lady Penelope‟s car does not break down and

Parker is sober. Her car won‟t break down unless Parker is drunk. But if he were

drunk, he would not drive. So, if Parker is driving, he is sober and Brains‟ plan will

work.

(W = Brain‟s plan will work; B = Lady Penelope‟s car will not break down;

S = Parker is sober; D = Parker will drive)

Making assumptions in the shortcut method.

In a few, exceptional cases, we arrive at situations where we are not forced to make

assignments but still do not have enough information to determine whether the argument is

valid or not. In such situations, we have to consider certain possible truth values in order to

get our answer.

Example: Determine whether or not the following SL-argument is valid:

P Q, P Q, Q P ╞═ P & Q

In this situation, no further

assignment is forced on us. But, we

know that „P‟ is either True or False.

If we can show that we get a

contradiction both on the assumption that it is true and on the assumption that it is false, then

P Q, P Q, Q P ╞═ P & Q

T T T F

1 2 3 4


we will have shown that it is not possible for the premises to be true and the conclusion to

befalse. So, we might set things out as follows.

Case 1: „P‟ is True.

Steps 9 and 8 contradict step 3. The

„@‟ beneath the final „P‟ signifies

that the assigned value is an

assumption. The number in

brackets under step 9 tells us

where the value was derived from. I could, for example, have determined the value under

the negation to be F, on the basis of steps 3 and 8. But in that case, step 9 would have

contradicted step 5.

The contradiction we have discovered tells us that the premises cannot be true while the

conclusion is false when „P‟ is True. Having ruled out this possibility, we now need to

consider the possibility that „P‟ is False.

Case 2: „P‟ is False.

Steps 7 and 9 are contradictory. In

other words, the premises cannot

be true while the conclusion is false

when „P‟ is False. Since „P‟ has to

be true or false, we have shown that it is impossible for the premises to be true while the

conclusion is false, period. Thus, the argument is VALID.

Another example: Determine whether or not the following SL-argument is valid:

A B, B A ╞═ A B.

First step: assign the premises True and the

conclusion False.

Unfortunately, nothing is forced on us.

So now we must consider different possibilities. Take „A‟, for instance. It has just two possible

truth values, True and False. We can check whether anything is forced on us by either

assumption.

Case 1: „A‟ is True

There is no contradiction. When „A‟ is True and „B‟ is

False, the premises are true and the conclusion is

false. So the argument is INVALID.

Because we have shown that the premises can be true while the conclusion is false, we do

not need to consider the other alternative, that „A‟ is False. We have done enough to

prove the argument INVALID.

P Q, P Q, Q P ╞═ P & Q

F T F T T T F T F F

6 1 7 2 9 3 8 @ 4 5

(5)

P Q, P Q, Q P ╞═ P & Q

T T T T T T F T F F

5 1 7 6 2 8 9 3 @ 4

A B, B A ╞═ A B

T T F

1 2 3

A B, B A ╞═ A B

T T F T T T T F F

5 1 7 8 2 6 @ 3 4

(4)


More exercises on the shortcut method

Determine whether the following arguments of SL are valid or invalid using the shortcut

method:

1. (A B) ╞═ A & B

2. P Q ╞═ (P & Q) & ( P & Q)

3. (L M) & (P M) ╞═ P L

4. M A, C H, A X, C M ╞═ X

5. L (A Q), L A, (L & Q) |= L Q

Derivations in Sentential Logic (1):

Some Simple Derivation Rules

Introduction

In the first half of the course we considered the logical language Sentential Logic (SL),

including the truth-table definitions of its connectives and accompanying account of SL-

validity. One motivation for such a language was to provide a simple framework for

determining the validity of natural-language arguments. In this section and the next, we will

consider a language which has precisely the same sentences as SL—so we shall continue to

refer to it as „SL‟—but where the connectives are defined by basic derivation (or inference)

rules. We can then prove that one sentence (the conclusion) is derivable from other sentences

(the premises) by showing that we can reach the conclusion from the premises by way of

simple steps sanctioned by the basic derivation rules.

Here is an example of a derivation (so you know what one looks like):

DERIVATION EXAMPLE 1

This is a derivation of „L & G‟ from the three

premises „(F G) H‟, „M & G‟ and „H

L‟. The numbers in the left-most column

enable one to refer to steps at various parts of

the argument. So e.g. the right-most column,

the justification column, at step 4 tells us that

step 4 was derived from step 2, using

derivation rule „&E‟ (to be explained shortly).

The vertical line is called a scope line—why we need it will become clearer in the next section.

The phrase „Prem.‟ in the justification column indicates that the sentence on that line was

given as a premise. Let us now consider the basic derivation rules used in the derivation.

Some basic derivation rules

There are two kinds of basic derivation rules, Introduction and Elimination Rules, for each

connective, which we will signify by suffixing the connective with „I‟ or „E‟, respectively. Here

are the rules we used in DERIVATION EXAMPLE 1.

1.

2.

3.

4.

5.

6.

7.

8.

(F G) H

M & G

H L

G

F G

H

L

L & G

Prem.

Prem.

Prem.

2, &E

4, I

1, 5, E

3, 6, E

4, 7, &I


m P

n Q

P & Q m, n &I

Q & P m, n, &I

&-Introduction (&I)

If sentences P and Q appear as steps in a derivation,

one may derive P & Q or Q & P. In the

justification column, one puts down the line-

numbers of the steps where P occurs and where Q

occurs and the name of the rule, „&I‟.

m P & Q

P m, &E

Q m, &E

&-Elimination (&E)

If P & Q appears as a step in a derivation, one may

derive P or Q. In the justification column, one puts

down the line-number of the line where P & Q

appears and the name of the rule, „&E‟.

Step 4 of DERIVATION EXAMPLE 1 uses the &-Elimination rule: given this rule, step 4

follows from step 2.

Step 8 of the derivation uses the &-Introduction rule: given this rule, step 8 follows from

steps 4 and 7.

Here are the other rules used in DERIVATION EXAMPLE 1:

m P

P Q m, I

Q P m, I

-Introduction (I)

If P appears as a step in a derivation, one may

derive P Q or Q P. In the justification column,

one puts down the line-number of the line where P

appears and the name of the rule, „I‟.

m P Q

n P

Q m, n E

or

m P Q

n Q

P m, n E

-Elimination (E)

If sentences P Q and P appear as steps in a

derivation, one may derive Q. If sentences P Q

and Q appear as steps in a derivation, one may

derive P. In the justification column one puts down

the line numbers of the steps where P Q occurs

and where P (or Q) occurs and the name of the

rule, „E‟.


m P Q

n P

Q m, n E

-Elimination (E)


derivation, one may derive Q. In the justific-ation

column, one puts down the line-numbers of the

steps where P Q occurs and where P occurs and

the name of the rule, „E‟.

The -Elimination rule was used in step 7: from „H L‟ (step 3) and „ H‟ (step 6), one

may infer „L‟. Note: you need both steps 3 and 6 in order to derive ‘L’ (it does not follow

from step 3 by itself).

The -Elimination rule is used in step 6 of the derivation: the rule sanctions the derivation

of step 6, „ H‟, from steps 1 and 5, „(F G) H‟ and „F G‟. (Again, you need both

steps 1 and 5 to derive step 6.)

We will consider two more rules in this section: -Elimination and -Elimination.

m P Q

n P

Q m, n E

or

m P Q

n Q

P m, n E

-Elimination (E)


derivation, one may derive Q. If sentences P Q

and Q appear as steps in a derivation, one may

derive P. In the justification column one puts down

the line numbers of the steps where P Q occurs

and where P (or Q) occurs and the name of the

rule, „E‟.

m P

P m, -E

-Elimination (E)

If sentence P appears as a step in a derivation,

one may derive P. In the justification column, one

puts down the line-numbers of the step where P

occurs and the name of the rule, „E‟.


Examples. Using the rules we have considered so far, construct derivations for the following

arguments („├—‟ is the derivability sign; it separates the premises from the conclusion):

(i) A B, C & B ├— A & C

1

2

3

4

5

6

7

A B

C & B

C

B

A

A

A & C

Prem.

Prem.

2, &E

2, &E

1,4 E

5, E

3,6 &I

(ii) (M Q) L, L X, Q ├— X

1

2

3

4

5

6

(M Q) L

L X

Q

M Q

L

X

Prem.

Prem.

Prem.

3, I

1,4 E

2,5 E

Exercises

Complete the derivations below by entering the appropriate justifications:

(a)

1 A (L M)

2 L & A

3 L

4 A

5 L M

6 L

7 M

8 A K

9 M & (A K)

Prem.

Prem.

_________________

_________________

_________________

_________________

_________________

_________________

_________________

1 R (P Q)

2 X Q

3 M & X

4 X

5 Q

6 P Q

7 R

8 M

9 M & R

Prem.

Prem.

Prem.

______________

______________

______________

______________

______________

______________


Strategies for constructing derivations

If an SL-argument is valid, then there will be a derivation for it; conversely, if the

argument‟s conclusion is derivable from its premises, then it will also be a valid argument. But,

unlike the truth-table or shortcut method, there is no sure-fire recipe for constructing

derivations. Failure to construct a derivation may simply be due to our failing to see how it

should be done. It is only by attempting (and thinking hard) about derivations that we can get

better at constructing them.

One way of approaching the construction of a derivation is to work from the bottom up.

Here is an example. Suppose we are asked to construct a derivation for the following

argument (from R.L. Simpson, Essentials of Symbolic Logic, p. 71):

B, B A, (A & B) R ├— R.

FIRST STAGE

Sketch the initial shape of the derivation. We don‟t know yet

how many steps will be needed, so, for the time being, we

use „z‟ in place of the last line-number.

Now, we ask ourselves: how could we have got „R‟ from

the premises? Evidently, only by using step 3. „R‟ does not

follow from step 3 by itself; but we know that if we had „A

& B‟ as a step in our derivation, then our rule -Elimination would permit us to derive „R‟.

So our derivation should have this shape:

SECOND STAGE

We put down „A & B‟ as the previous step and number the

line „y‟. We now know how step z is to be justified: it

follows from steps 3 and y, using E.

Now, how could „A & B‟ be derived? It does appear as the

antecedent of the conditional in step 3, but no rule allows us

to derive the left-hand-side of a conditional. So, we must

look elsewhere. There is only on way we could have got „A

& B‟ in this derivation—by &-Introduction; that is, we must be able to get „A‟ and „B‟ in

earlier steps and used &-Introduction on these:

THIRD STAGE

We already have „B‟ as a step (it is the first premise), so,

prior to deriving „A & B‟, we must have derived „A‟. We

know now that the derivation must look something like the

figure on the left.

Next question: how could we have got „A‟? It could only

have been derived from the „ A‟ which appears as the

right-hand-side of the bi-conditional in step 2, „B A‟:

we know that -Elimination allows us to derive „A‟ from

„ A‟. See next diagram.

FOURTH STAGE

1

2

3

Z

B

B A

(A & B) R

R

Prem.

Prem.

Prem.

?

1

2

3

y

Z

B

B A

(A & B) R

A & B

R

Prem.

Prem.

Prem.

?

3,y E

1

2

3

X

Y

Z

B

B A

(A & B) R

A

A & B

R

Prem.

Prem.

Prem.

?

1,X &I

3,y E


Next: how could „ A‟ be derived? It does not follow

from step 2 by itself. Our rule for -Elimination tells us

that we can derive one side of a bi-conditional (double

arrow) only if we have the other side as a step. But we do:

namely, „B‟ in step 1. Hence, we do not need any further

intermediate step to derive „ A‟— it follows from steps 1

and 2, using -Elimination. The justification for step w is

simply: „1,2 E‟.

FINAL STAGE

Remember to replace the letters used in place of line-

numbers with numbers, and to alter the justifications in the

right-hand column accordingly. (There is no need to show

the various stages of your construction—just give the final,

complete derivation as in the diagram)

A useful tip if you are stuck: use any elimination rules you can! This is a „pot-luck‟

method and may not always work, but often it will indicate how the remainder of the

derivation could be constructed. Here is an example. Construct a derivation for the following

argument: M & R, (Q M) R ├— Q W.

STAGE 1

Write down the premises and, to remind you of

what it is you are trying to derive, the conclusion.

STAGE 2

We know that if an &-statement is a step, we are

allowed, by &-elimination, to derive both sides—so

do so, and see if any other eliminations suggest

themselves. As it turns out, we can use E and

E.

1

2

3

W

X

y

Z

B

B A

(A & B) R

A

A

A & B

R

Prem.

Prem.

Prem.

?

1,W E

1,X &I

3,y E

1

2

3

4

5

6

7

B

B A

(A & B) R

A

A

A & B

R

Prem.

Prem.

Prem.

1,2 E

4, E

1,6 &I

3,6 E

1

2

z

M & R

(Q M) R

Q W

Prem.

Prem.

?

1

2

3

4

z

M & R

(Q M) R

M

R

Q W

Prem.

Prem.

1, &E

1, &E

?


STAGE 3

Step 5 follows from step 3, using -Elimination;

and Step 6, follows from steps 2 and 4. NOTE: you

need both steps 2 and 4 to derive step 6.

STAGE 4 (FINAL ANSWER)

Step 6 is the disjunction „Q M‟. We can derive

either side of the disjunct if we have the explicit

negation of the other side as a step (this is what the

-Elimination rule tells us). Thus, we may derive

„Q‟ from steps 3 and 6. NOTE: it would be incorrect

to derive it from steps 5 and 6; step 5, because the

explicit negation of „ M‟ is not „M‟ but „ M‟.

Once we have got „Q‟ (step 6), we may derive „Q

W‟, using -Introduction (even though „W‟

appears nowhere else in the derivation!). So, the

derivation is complete.

Exercises

1. Locate the errors in the derivation below (explain why they are errors):

2. Construct derivations for each of the following arguments using any of the rules we have

considered in this section:

(a) K & ( L K) ├─ L

(b) R M, (R K) Z, R ├─ Z & M

(c) P R, R & X, Y P ├─ Y

(d) P (M Z), (A P) & (M Z) ├─ A

(e) (F & G) H, H F, (F & G) ├─ F G

1

2

3

4

5

6

z

M & R

(Q M) R

M

R

M

Q M

Q W

Prem.

Prem.

1, & E

1, &E

3, E

2,4 E

?

1 M & R Prem.

2 (Q M) R Prem.

3 M 1 &E

4 R 1 &E

5 M 3 E

6 Q M 2,4 E

7 Q 3,6 E

8 Q W 7 I

1

2

3

4

5

6

A

B (A C)

A C

B

C

C B

Prem.

Prem.

2, E

2, E

3 E

5, I


Derivations in Sentential Logic (2): Assumptions

Assumptions

Some derivations require one to make assumptions at some stage. For example, in some

cases, the best (or only!) way of showing that a certain proposition, p, is false is to show that

an absurd or contradictory result follows if one assumes the truth of p. Conversely, we could

show a proposition to be true by showing that an absurd or contradictory result follows from

the assumption that it is false.

In fact, this is precisely the idea behind the shortcut method we used in Sentential Logic.

The method begins by assuming the premises to be true and the conclusion to be false; if this

results in a contradiction, we concluded that the argument was valid—that the conclusion had

to be true if the premisses were.

Reductio ad absurdum: this is the name given to the style of argument just mentioned, i.e.

an argument for a conclusion by way of show that assuming otherwise leads to absurdity or

contradiction. The phrase indirect proof is also used, though generally restricted to cases

when one argues for a conclusion by way of showing that assuming otherwise leads to

contradiction.

Two of our derivation rules for Sentential Logic, -Introduction and -Introduction,

proceed from assumptions.

-Introduction.

Let us approach this rule by asking: when are we entitled to infer an arrow-sentence of the

form „P Q‟? Earlier, we took such a sentence to affirm something like: If P is true, then so

is Q. Our rule for arrow-introduction (below) basically says that if on the assumption that P is

true we can derive Q using any of the derivation rules, then we are entitled to derive „P Q‟.

m P Ass.

n Q

P Q m, n I

-Introduction (I)

If one can derive Q by assuming (or under the

assumption of) P, then one may derive P Q. In

the justification column, one puts down the line

numbers of the steps at which P is assumed and at

which Q is derived, and the name of the rule, „I‟.

Here is an example of a derivation using this rule. To show:

M P, P Q |— M Q

1. M P Prem.

2. P Q Prem.

3. M Ass.

4. P 1,3 E

5. Q 2,4 E

6. M Q 3,5 I

Explanation

In order to prove the arrow-statement „

M Q‟, we assume the (whole) left-hand

side of the arrow (in this case, „ M‟) and

derive the right-hand side (in this case

„Q‟).


Whenever we make an assumption in a derivation we introduce another scope line (moving

over one place to the right, as it were). This is what we have done at step 3 above. We do so

because we must distinguish between what follows from our original premises and what

follows from them given that the assumption is also true. When we discharge (finished with)

the assumption, we move back one scope line.

Another example. Construct a derivation for:

A, (A & B) Q |— (B & C) (C & Q)

1 A Prem. Explanation.

In order to prove an arrow-statement (see step

10), we assume the left-hand side (the

antecedent) of the arrow statement and show

that the right-hand side (the consequent) is

derivable. In this proof the antecedent has been

assumed at step 4. Notice that we have added

another scope line a bit to the right of the main

one—to indicate that these steps are what

follow given the assumption. The consequent

has been shown to be derivable under the

assumption at step 9. This entitles us to derive

the arrow statement at step 10.

2 (A & B) Q Prem.

3 A 1, E

4 B & C Ass.

5 B 4, &E

6 A & B 3,5 &I

7 Q 2,6

8 C 4, &E

9 C & Q 7,8 &I

10 (B & C) (C & Q) 4,9 I

Exercises

Construct derivations for each of the following arguments using any of the rules we

have considered so far:

(1) (P L) M, K M |— L ( M &

(2) ( X & Y) Z, Y X |— (H & Y) Z

(3) L, (M X) L |— (L M) X

(4) A (B C) |— (A & B) (C Z)

-Introduction. This rule just reflects the point made earlier that one can establish a

conclusion by showing that assuming otherwise leads to a contradiction.

m P Ass.

n Q

o Q

P m, n, o I

-Introduction (I)

If one can derive a sentence, Q, and its explicit

negation, Q, assuming P, then one may derive

P. In the justification column, one puts down the

line numbers of the steps at which P is assumed,

and at which Q and Q occur, and the name of the

rule, „I‟.


Here‟s an example using this rule, a derivation for: F (G & H) |— (H & F)

1 F (G & H) Prem. Explanation.

We have assumed the precise „opposite‟

of the conclusion at step 2. At steps 5

and 6 we have derived a statement and

its explicit negation. So we are entitled

to conclude that the original assumption

was false: i.e. we can derive the

negation of the assumption.

2 H & F Ass.

3 F 2, &E

4 G & H 1,3 E

5 H 4, &E

6 2, &E

7 (H & F)

Another example. A derivation for: (A B), (A B) D |— D & A

1 (A B) Prem.

2 (A B) D Prem.

3 D 1,2 E

4 A Ass.

5 4, I

6 1, R

7 A

8 D & A

Step 6 uses a new, but quite trivial, rule, Reiteration. This rule basically allows one to

reiterate a step later in the same derivation or any sub-derivation. We need it in this case

because our rule for -Introduction requires us to have the contradictory steps (P and P)

appearing under the assumption (that is, in the same scope line as the assumption).

m P

P m, R

m P

P m, R

Reiteration (R)

If sentence P appears as a step in

a derivation, one may derive P

later in the derivation or in any

sub-derivation.


Exercises

Construct derivations for each of the following arguments using any of the

rules we have considered so far:

(1) P & Q |— P Q)

(2) (F G) M, F M |— F

(3) A B) C |— (C & A) B

(4) G (H & Y), ( F K) G, F ├— K Y

(5) J ( J & D), D F, F J ├— J

The power of -Introduction

Derivation (5) in Exercises 2 above reveals how -introduction may be used to derive

conclusions that are NOT negation-statements. For, from “ J” we may derive “J” by use of

the -Elimination rule. Thus, we can construct a derivation for the following argument:

J ( J & D), D F, F J ├— J

namely, by deriving “ J” as we did for (5) by using -Introduction, and then deriving “J” by

using -Elimination. In fact, we cannot derive “J” from the above premises unless we use

-Introduction.

Double assumptions

A derivation may involve two or more assumptions, one after the other. Consider e.g.

(P Q) |— P & Q

To derive the conclusion one has to derive “ P” by way of -Introduction, then derive “

Q” by way of -Introduction, and, finally derive the conclusion by way of &-Introduction.

Exercises: Construct a derivation for the argument above. Then construct a derivation for:

P & Q |— (P Q)

This does not require two assumptions, but by constructing both derivations you have shown

that “ (P Q)” and “ P & Q” are logically equivalent!

Introduction to Predicate Logic

The inadequacy of Sentential Logic

The shortcut method provides a quick and effective way of determining whether an

argument of Sentential Logic (SL) is valid or not. But there are many arguments of

ordinary language where it seems the translation into SL is inadequate. Consider e.g. the

following argument we have come across:

Premises:

Conclusion:

All vegetarians are healthy.

Tracy is a vegetarian.

Hence, Tracy is healthy. Valid


Because none of the statements involved contain sentence connectives, each of them must be

treated as simple (i.e. atomic) sentences of SL. So, we get something like the following SL-

invalid argument on translation: V, T ╞═ H.

Another example:

Premises:

Conclusion:

Superman can fly.

Batman cannot fly.

Hence, Superman is not Batman. Valid

This argument would get translated into SL as: S, B |= I (S = Superman can fly; B =

Batman can fly; I = Superman is Batman). This too comes out invalid using the shortcut

method.

What gives? The problem is that we need more than sentence connectives to capture the

„shape‟ of sentences. In the latter argument, for example, we need to capture the fact that one

and the same property (of being into fly) is invoked in the two premises; in the first argument,

we need to capture the properties of being a vegetarian and being healthy. So, we need a

more sophisticated logical language that breaks sentences down into simpler components.

Enter, Predicate Logic.

Predicate Logic (PL):

Constants, Predicates and Quantifiers

VOCABULARY CATEGORY USE

a, b, c, ..., w Constants / Names Designating objects

A, B, C, ..., Z Predicate letters Ascribing properties

x, y, z Variables Quantifying over objects

(x) Existential Quantifier There is an x such that ...

(x) Universal Quantifier Every x is such that …

, &, , , Logical connectives Same as Sentential Logic

Constants (names) are used when a particular thing is being referred to, and the

predicate letters are used to ascribe properties to that thing. So, e.g., we may symbolize

Harry is bald as „Bh‟, where „Bx‟ stands for „x is bald‟ and „h‟ stands for Harry. If Harry is

dull, so is Betty may be translated into Predicate Logic (PL) as: Dh Db (where „Dx‟ stands

for „x is dull‟ and „b‟ for Betty).

NOTE: it would be wrong to translate Politicians are dull as „Dp‟ (with „p‟ standing

for politicians). For this statement is not about a particular individual but about all individuals

with a certain property. (We‟ll see how to symbolize this kind of statement next term.)

Predicate letters are also used to signify relations between individuals. E.g. „Txy‟ could

stand for „x is taller than y‟; „Bxyz‟ for „x is between y and z‟. (Note: the former is called a

two-place predicate, because it has two „place-holders‟; „Bxyz‟ is a three-place predicate, and

so on.)


An n-place predicate letter followed by n names (constants) is called an atomic sentence of

PL. One can construct complex sentences from atomic sentences with the logical connectives

of Sentential Logic, the translations of the connectives being the same as before.

SENTENCE TRANSLATION

I am taller than Jenny. Tmj (m = me)

Kevin is not older than Marion. Okm

Jenny is taller than both Kevin and Marion. Tjk & Tjm

Marion is older than Kev if Jenny is. Ojk Omk

Brighton is not between Oxford and Glasgow. Bbog

Exercises

1. Translate the following sentences of Predicate Logic into English:

(a) Fba & Fma (Fxy = x is y‟s friend; b = Ben; a =Alice; m = me)

(b) (Tma & Tbm) Tba (Txy = x is taller than y)

(c) Sp Ad

(Sx: x will survive; Ax: x arrives soon; p = the patient; d = the doctor)

(d) (Sb Sa) & (Sb & Sa)

2. Translate the following English sentences into Predicate Logic (using the above key:

(a) Tracy is a vegetarian if her brother is.

(Vx = x is vegetarian; t = Tracy; b = Tracy‟s brother)

(b) Neither Tracy nor Alice are friends of mine. (Use previous definitions)

(c) Ben won‟t survive unless Alice arrives soon. (Use previous definitions)

(d) I‟ll be Alice‟s friend only if she isn‟t Tracy‟s friend.

(e) Alice and Tracy are the same height. (Using Txy = x is taller than y).

(f) I‟ll be in trouble if Tracy believes either Ben or Alice.

(Tx = x is in trouble; Bxy = x believes y)

Existential quantifiers are used to affirm the existence of objects possessing certain

properties. „(x)Fx‟ means:

There is an individual (or object) which is (an) F.

So e.g. we may translate „Unicorns exist‟ or „There are unicorns‟ into PL as „(x)Ux‟, where

„Ux‟ means „x is a unicorn‟.

NOTE: We could just as well have used: „(y)Fy‟ or „(z)Fz‟. The variables „x‟, „y‟, and

„z‟ do not stand for any specific individuals. So „Fx‟, for example, is not a grammatical

sentence. Neither is „(y)Fy & Gy‟, because the „y‟ in „ Gy‟ is not bound (governed) by

the quantifier; „(y)(Fy and Gy)‟, on the other hand, is fine: both tokens of „y‟ are bound by

the existential quantifier. Variables can only occur in sentences if they are bound (governed)

by a quantifier.

NOTE: PL does not discriminate between the following claims:

There are Fs; there is at least one F; there are many Fs; something is F

All x)Fx‟.


Statements of the form, „Nothing is F‟ simply deny that something is F. So they may be

translated as „ (x)Fx‟.

Statements of the form, „Some Fs are G‟ or „Some F is G‟ are logically equivalent to

„Something is both F and G‟. So they may be translated as „(y)(Fy & Gy)‟.

„No F is G‟ simply denies that some F is G. So it may be translated as „ (z)(Fz & Gz)‟.

Examples


There are many dogs. (x)Dx

Absolute cads exist! (y)Ay

Someone is not an absolute cad. (x) Ax

There are no dragons. (y)Dy

If someone is drunk, Alan is. (x)Dx Da

Something is bothering Alice. (x)Bxa

Nothing bothers me. (x)Bxm

Some cats are black. (z)(Cz &Bz)

Some cats are not black. (z)(Cz & Bz)

A dog is chasing Harry. (x)(Dx & Cxh)

Martha is chasing someone. (x)Cmx

Martha isn‟t chasing anyone. (x)Cmx

No dogs chase Alice. (x)(Dx & Cxa)

Translation Exercises involving existential quantification

Use the following key for questions 1 and 2:

KEY

Cx : x is crazy a : Archie

Ex : x is an elephant b : Blanche

Sx : x smokes m : Me

Vx : x is a vegetarian t : that room

Ixy : x is in y u : this university

Mxy : x is married to y

1. Translate the following PL-sentences into natural English (do your best with (i) and (j)!):

a. Iat & Iau

b. Mab (Sa Sb)

c. (y)Iyt

d. (x)(Mxb & Vx)

e. (z)(Ez & Cz)

f. (y)Mya Em

g. (y)Mby (

h. (x)(Ixt & Sx)


i. (y)((Cy & Ey) & Iyu)

j. (x) (y)Mxy

2. Translate the following sentences into sentences of Predicate Logic using existential

quantifiers (where necessary!):

a. I‟m not married to Archie.

b. Blanche is a vegetarian only if she is crazy.

c. Nobody‟s crazy.

d. I don‟t smoke, but many do.

e. Archie is married. (Clue: Archie is married if he‟s married to someone)

f. Some elephants are vegetarians.

g. Neither Blanche nor Archie are in this university.

h. Someone in that room is not a vegetarian.

i. No one is crazy if Archie isn‟t.

j. Unless I‟m crazy, there‟s an elephant in that room.

Predicate Logic: Universal Quantifiers and Domains Of Quantification

Domains of Quantification

Consider the following pairs of sentences:

(a1) Some students missed today‟s class.

(a2) Some students want to be academics.

(b1) Everyone enjoyed the party.

(b2) Everyone is to blame for the state of our society.

(a1) is presumably true if and only if there are students officially taking the course in

question that missed the class; so the fact that there are other students (e.g. students in other

universities) who did not attend the class does not automatically make (a1) true. (a2), on the

other hand, is presumably true if there are students somewhere—at any rate, not necessarily

those officially taking the course alluded to in (a1)—who want to be academics. So, the group

of individuals under discussion in (a1) is different from those under discussion in (a2). Thus, the

domain of quantification (or universe of discourse) is said to be different in the two cases.

Likewise, the quantifier „everyone‟ ranges over one set of individuals in (b1)—namely, the

people who came to the party—and a different set of individuals in (b2)—namely, the people

of this country (or the world). The domains of quantification are the respective sets of

individuals.

The upshot of all this is that whenever quantifiers are used in ordinary language, a domain

of quantification is generally assumed or taken for granted. When we translate into Predicate

Logic, however, we should, strictly speaking, make the domain explicit—especially when the

domain varies from question to question (or sentence to sentence). But in this course, we will

specify domains only when the context is ambiguous.

Universal quantifiers: (x), (x)

Universal quantifiers are used when we wish to affirm that everything (or everything of a

certain kind) possesses a certain property. „(x)Fx‟ means: For each and every x, x is F. So


e.g. we may translate „Everyone is moral‟ as „(x)Mx‟, where „Mx‟ means „x is moral‟;

„Everyone likes Archie‟ may be translated as „(y)Lya‟; and so on.

Note: the following statements are equivalent:

Every F is G; all Fs are G; whatever (whoever) is F is G

A statement of the form, „No one is (an) F‟ is equivalent to „Everyone is non-F‟ and can be

translated as „(x) Fx‟. (Ditto for: „Nothing is F‟).

Warning!!: „Everyone is not a smoker‟ (or „all are not smokers‟) generally does not mean:

„everyone is a non-smoker‟. Rather, it is more likely to be used to deny that everyone is a

smoker. So it should be translated as: (x)Sx. (Ditto for other statements of the same form.)

Universal quantifiers are also used to affirm that everything with a certain property has

another property. But it would be incorrect to translate All dogs bark, for example, as:

(x)(Dx & Bx). This PL-sentence affirms that everything is a barking dog! The correct

translation of „Every dog barks‟ (or „All dogs bark‟) is of the form: (x)(Dx Bx). Here‟s

why.

Suppose you know that every dog barks. What would you thereby know about any

randomly selected object from the world? You would know simply that: if the chosen object is

a dog, then it barks. So, it follows from the claim that every dog barks, that for any object, x,

if x is a dog, x barks; symbolically: (x)(Dx Bx).

Next, suppose it is true of each and every object in the world that: if it is a dog, it barks.

Then, clearly, it must be true that any given dog barks; in other words, it must be true that

every dog barks.

Hence, the correct translation of „Every dog barks‟ is: (x)(Dx Bx). More generally,

statements of the form: every F is G, any F is G, each F is G, all Fs are G, whatever is F is G

are all translated as (x)(Fx Gx).

Statements of the form: Only Fs are G are equivalent to Every G is F. So the PL-

translation is: (x)(Gx Fx)

We saw earlier that statements of the form No Fs are G can be translated as:

(x)(Fx & Gx). But this is equivalent to: (x)(Fx Gx). Either translation will do.


Everyone is hungry (y)Hy

Nothing is sacred (x) Sx

Every journalist drinks (x)(Jx Dx)

All yacht-owners love the Pacific Ocean (z)(Yz Lzp)

(key: Yx = x is a yacht-owner; Lxy = x loves y; p = the Pacific Ocean)

Any cow eats grass (y)(Cy Ey)

All bricklayers are not sexist (x)(Bx Sx)

(see earlier warning!)

No horse reads (x)(Hx Rx)

OR (x)(Hx & Rx)

Only horses run at Ascot (z)(Rz Hz)

Whoever hit Alan is vicious (y)(Hya Vy)


Translation Exercises

1. Using the key below, translate the following PL-sentences into English:

KEY

Cx : x supports City b : Bill

Lx : x likes football e : Eric

Sx : x is a snob m : Martina

Tx : x likes tennis p : Penny

Ux : x supports United

Bxy : x bores y Fxy : x is a fan of y

a. Lb Fbe

b. (x)Tx & (y)Ly

c. (y)(Fym & Bym)

d. (z)Fbz Fbb

e. (x)(Cx Bxm)

f. (y) Byp

g. (z)(Uz Cz)

h. (x)(y)Byx

i. (x)((Tx & Lx) Sx)

2. Using the following key, translate the following sentences into Predicate Logic:

KEY

Cx : x is crazy a : Archie

Ex : x is an elephant b : Blanche

Sx : x smokes m : Me

Vx : x is a vegetarian t : that room

Ixy : x is in y u : this university

Mxy : x is married to y

a. Everyone is crazy.

b. All elephants are vegetarians.

c. No elephant smokes.

d. Whoever Blanche is married to is a vegetarian.

e. Everyone in that room is a non-smoker.

f. Not everyone in this university is in that room.

g. Every vegetarian smoker is crazy.

h. Only elephants are in that room.

i. Everyone in that room is not a vegetarian.

j. If some elephants are vegetarians all are.

k. Someone is in this university if anyone is in that room.

l. If anyone is in that room, they are in this university.


Predicate Logic with Identity You may find the following useful for this section: S. Guttenplan, The Languages of Logic,

pp. 196-204 (Library code: [BC 71 Gut])

Adding a two-place identity predicate, „x ═ y‟ (read: x is identical with y) to Predicate

Logic (PL) increases the language‟s expressive power considerably. To begin with, we need

the predicate to express the identity or distinctness of individuals. E.g. “Superman is Clark

Kent” might be expressed as “s ═ c”, and “Bruce Wayne is not Robin” might be expressed as

as “b r” (where “x y” just abbreviates “ (x = y)”). But there are other significant uses.

Here are some examples.

(1) Carla didn’t go to the party but everyone else did.

How should (1) be expressed in Predicate Logic? (Take „Px‟ to stand for „x went to the party‟

and „c‟ for Carla). Here is a wrong answer:

(1WRONG) Pc & (x)Px

(1WRONG) affirms that: Carla didn’t go to the party and (but) everyone went. This is wrong

because it is contradictory: if Carla didn‟t go to the party, it cannot be true that everyone

went! Sentence (1), however, clearly is not contradictory. What we want to express is the

proposition that: Carla did not go to the party but everyone other than (or: everyone who is

not) Carla did go. This is captured by

(1RIGHT) Pc & (x)(x c Px)

(1RIGHT) also captures the content of “Everyone but [except, other than] Carla went to the

party” (though I would also accept „(x)(x c Px)‟ as a translation of “Everyone other than

Carla went to the party”).

(2) Only Phil drinks milk (No one but Phil drinks milk)

(Key: p = Phil, Mx = x drinks milk)

To see how (2) should be translated, it is useful to compare it with something like: Only cats

drink milk. This, we have seen, affirms the same thing as (M)

(M) Everything which drinks milk is a cat

(x)(Mx Cx)

So, (2) should be read as affirming that everything which drinks milk is (identical with) Phil:

(2a) (x)(Mx x = p)

NOTE: Just as (M) does not entail that there are cats who drink milk, (2a) does not entail that

Phil drinks milk. If this seems inadequate, i.e. if one does take (2) to entail that Phil drinks

milk, then (2) may be expressed as (2b) or (2c):

(2b) Mp & (x)(Mx x = p)

(2c) (x)(Mx x = p)

[It is the double-arrow in (2c) which ensures that if something is Phil, it drinks milk—i.e. that

Phil drinks milk.]


Another example: “Only Phil and Sue drink milk” is translated as:

(3a) (x)(Mx (x = p x = s))

(3a) is in fact compatible with neither Phil nor Sue being milk-drinkers. If one took the original

sentence to imply that they are, then one should translate it with the double-arrow, i.e. as:

(3b) (x)(Mx (x = p x = s))

(3) There are at least two Latvians (There is more than one Latvian)

(Key: Lx = x is a Latvian)

Here is a wrong translation of (3):

(3WRONG) (x)(y)(Lx & Ly)

The occurrence of an existential quantifier in a sentence simply signifies the existence of a

certain kind of thing: that there is at least one thing of that kind. Thus, (3WRONG) just affirms the

existence of Latvians „twice over‟, as it were: There is at least one Latvian and there is at

least one Latvian. To express the fact that there is more than one, we need to specify that

there are Latvians x and y which are distinct:

(3RIGHT) (x)(y)((Lx & Ly) & x y)

One would express the claim that there are at least four Latvians in the following way:

(x1)(x2)(x3)(x4)(Lx1 & … & Lx4 & x1 x2 & x1 x3 … & x3 x4)

(4) There is exactly one U.S. President (There is one, and only one, U.S. President;

there is only one U.S. President) (Key: Px = x is U.S. President)

(4) affirms that there is a U.S. President that s/he is the only U.S. President. This may be

expressed in Predicate Logic as either (4a) or (4b):

(4a) (x)(Px & (y)(Py x = y))

(4b) (x)(y)(Py x = y)

[Compare these with (2b) and (2c), respectively (on the previous page), which may be

regarded as expressing that Phil is the only (or the one and only) milk-drinker.]

Here is how one would symbolize “There are exactly (only) two tigers”:

(x)(y)((Tx & Ty & (z)(Tz (z = x z = y))) & x y)

Or: (x)(y)((z)(Tz (z = x z = y)) & x y)

(5) There is no more than one unicorn. (Ux = x is a unicorn)

At most one unicorn exists.

(5) does not affirm the existence of a unicorn—it does not state there is one unicorn but no

more. It simply denies that there are at least two unicorns. So one way of symbolizing it is

thus:

(5a) (x)(y)((Ux & Uy) & x y)


Another way of expressing the same fact is to maintain that if x and y are unicorns (for any x

and y), x and y are identical:

(5b) (x)(y)((Ux & Uy) x = y)

Exercises


a. (x)(x r Bxm)

b. (y)(Dy y ═ t) (t = Tibbles)

c. (y)(Byt y = r)

d. (x)(Dx & (y)(Dy & y x))

2. Translate the following sentences into Predicate Logic with Identity:

a. At least two dogs bit Tibbles. (Dx; Bxy; t)

b. Rover bit everyone but me. (Bxy; r; m)

c. Rover bit me, and so did someone else.

d. At most one dog bit me.

3. Give a simpler Predicate Logic translation of: (y)(Bym & y = t)

Exercises 4

Use the following key for questions 1 and 2 (the asterisked questions involve identity):

KEY

Ax : x is an actor (actress) a : Akira

Dx : x is a director b : Boris

Hx : x is happy d : Doris

Ox : x has won an Oscar j : Juanita

Bxy : x is a better actor than y m : Me

Fxy : x is a fan of y s : Sanjay

Mxy : x has met y

Wxy : x has worked with y


a. Fab Fas

b. (Wda Wdb) & Wdm

c. (y)(Ay & Hy) & (y)(Ay & Hy)

d. (z)(Wza Fmz)

e. (x)((Dx & Wxj) Ox)

f. (x)(Wdx & Bxd)

g. (x)(Ax & Dx) (y)(Ay Hy)

h. (x)(Dx (y)(Ay & Fyx))

i. (y)(Wby Oy)

*j. b=s (x)(y)((Fxm & Fym) x=y)


*k. (z)((Dz & Az) s=z)

*l (x)((Mbx & Fxb) & (y)(Fyb x=y))

2. Translate the following English sentences into Predicate Logic:

a. I‟ve met Akira—he is a director.

b. I won‟t be happy until I‟ve met Doris.

c. Anyone who has worked with Juanita is a fan of hers.

d. I‟ve met an Oscar-winner!

e. No one is a better actor than Boris.

f. Doris has a fan who is a better actress than her!

g. All Oscar-winning directors are happy.

h. If Sanjay is happy, I‟ve met a happy actor.

i. Boris has met an actress and a director she is working with.

j. If anyone is a better actor than Akira, Juanita is.

k. My fans will be unhappy only if I do not win the Oscar.

*l. Everyone but Sanjay has worked with Doris.

*m. I‟ve only worked with Boris and Akira.

*n. Sanjay is the only happy actor.

Leibniz’s Law(s)

The Indiscernibility of Identicals: a = b (Fa Fb)

This is a general principle concerning identity that is often simply called Leibniz’s Law. It

maintains that if a and b are identical, then whatever is true of a is true of b, and vice versa

(for any a and b). So e.g. if Clark Kent is Superman, then Clark Kent can fly if and only if

Superman can.

The Identity of Indiscernibles: (Fa Fb) a = b

This is another principle Leibniz held, but it is rather more contentious. It maintains that if

everything true of a is also true of b, and vice versa, then a is identical with b.


Predicate Logic Derivations: Two Simple Rules + Examples + Exercises

Predicate Logic Derivation Rules: -Introduction and -Elimination

m

(v)* m, I

-Introduction ()

From a PL-sentence containing occurrences of a

name, c, one may derive (v)*, where v is a

variable (e.g. x, y, or z) that does not appear in ,

and * is the result of replacing one or more (or

all) occurrences of c in with v.

m (v)

* m, E

-Elimination (E)

From a PL-sentence of the form „(v)‟, one may

derive *, where * is the result of replacing every

occurrence of v in with a particular name, c.

Some sample derivations: please complete the justification columns

1. (x)(Fx & Gx), Ft Ht |— Ht

1 (x)(Fx & Gx) Prem

2 Ft Ht Prem

3 Ft & Gt

4 Ft

5 Ht

2. Fb, (x)(Fx Gx) |— (z)(Fz & Gz)

1 Fb Prem

2 (x)(Fx Gx) Prem

3 Fb Gb

4 Gb

5 Fb & Gb

6 (z)(Fz & Gz)


3. (x)Fxb (x) Fxb, Fab |— Fhb

1 (x)Fxb (x) Fxb Prem

2 Fab Prem

3 (x)Fxb Ass.

4 Fab

5 Fab

6 (x)Fxb

7 (x) Fxb

8 Fhb

4. Ft Gt, (z)(Gz Hz) (x)Gx |— Ft Gm

1 Ft Gt Prem

2 z)(Gz Hz) (x)Gx Prem

3 Ft

4 Gt

5 Gt Ht

6 (z)(Gz Hz)

7 (x)Gx

8 Gm

9 Ft Gm

Construct derivations for the following PL-arguments:

5. (x)(Mx Hx), (y) Hy |— Mk & Hk

6. Abm & Fmb, (x)Abx (x)Fbx |— (y)(Fby & Fyb)

(Clue: derive “Fbm & Fmb” before the conclusion)


Predicate Logic Derivations: Two More Rules + Examples + Exercises

Predicate Logic Derivation Rules: -Introduction and -Elimination

m

(v)* m I

-Introduction ()

From a PL-sentence containing occurrences of a

name, c, that does not appear in the premises or

any governing assumptions, one may derive (v)*,

where v is a variable (e.g. x, y, or z) that does not

appear in , and * is the result of replacing every

occurrence of c in with v.

m (v)

n * Ass.

o α

α m,n,o E

-Elimination (E)

One may derive a statement α from (v) if one can

derive α under the assumption of *, where * is the

result of replacing every occurrence of v in with a

name, c, that does not appear anywhere earlier in

the derivation or in α.

Some sample derivations

1. (x)(Fx & Gx) |— (y)Gy

1 (x)(Fx & Gx) Prem

2 Fa & Ga 1 E

3 Ga 2 &E

4 (y)Gy 3 I

2. (x)(Vx Hx), (x)(Hx Tx) |— (x)(Vx Tx)

1 (x)(Vx Hx) Prem

2 (x)(Hx Tx) Prem

3 Vb Ass.

4 Vb Hb 1 E

5 Hb Tb 2 E

6 Hb 3,4 E

7 Tb 5,6 E

8 Vb Tb 3,7 I

9 (x)(Vx Tx) 8 I


3.y)Gy |— (z)Gz

1 y)Gy Prem

2 Gm Ass.

3 y)Gy 2 I

4 y)Gy 1 R

5 Gm 2,3,4 I

6 (z) Gz 5 I

4. Ft, (x)(Fx Gx)|— (z)(Fz Gz)

1 Ft Prem

2 (x)(Fx Gx) Prem

3 Ft Gt 2 E

4 Gt 1,3 E

5 Ft & Gt 1,4 &I

6 (z)(Fz Gz) 5 I

5. (x)(Txa Txb), (x)xa |— (x)Txb

1 (x)(Txa Txb) Prem

2 (x)xa Prem

3 Tka Tkb Ass.

4 ka 2 E

5 Tkb 3,4 E

6 (x)Txb 5 I

7 (x)Txb 1,3,6 E

Exercises

Construct derivations for the following arguments of Predicate Logic:

1. (y)Ay, (z)Bz |— (x)(Ax & Bx)

2. (y)Cy Cb |— (x)(Cx Cb)

3. (x)(Lx tx & Bx)), Lb & Db |— (y)Mty

4. Ft Gt, (z)Gz (x)Gz ├— Ft Gm

5. (x)(Fx Gx), (y)Fy ├— (x)(Fx & Gx)

6. (y)(Ty & Qay), (y)Qay Qaa├— (y)Qyy

End of course. Good luck in the exam!

Department of Philosophy, WITS PHIL 3009 SYMBOLIC … logic... · Department of Philosophy, WITS PHIL 3009 SYMBOLIC LOGIC COURSE PACK Lecturer: MURALI RAMACHANDRAN INTRODUCTION: DEDUCTIVE

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