Lesson 3 Validity of arguments Marie Duží. Logical entailment A formula A logically follows from a set of formulas M, denoted M |= A, iff A is true in.

LLesson 3 esson 3

Validity of argumentsValidity of arguments

Marie DužíMarie Duží

Logical entailment

• A formula A logically follows from a set of formulas M, denoted M |= A, iff A is true in every model of the set M.

• Recall Definition 1. The circumstances are modelled in accordance with the expressive power of a given logic.

• PL valuations (True – 1, False - 0) of elementary atomic sentences

• FOL interpretations of predicate and functional symbols

– “Under all the circumstances” means in all valuations / interpretations that make premises true (i.e. in all models of the premises) the conclusion must be true as well.

Logical entailment in PL• He is at home (h) or he has gone to a pub (p)• If he is at home (h) then he is waiting for us (w) If he is not waiting (w) for us then he has gone to the pub (p).

h, p, w | h p, h w | w p

1 1 1 1 1 1 conclusion 1 1 0 1 0 1

1 0 1 1 1 1 is true in all1 0 0 1 0 0

0 1 1 1 1 1 the four models 0 1 0 1 1 1 of premises

0 0 1 0 1 10 0 0 0 1 0

Logical entailment in PL• He is at home (h) or he has gone to a pub (p)• If he is at home (h) then he is waiting for us (w) If he is not waiting (w) for us then he has gone to the pub (p).

h p, h w | w p• The table has 2n lines!

Hence, an indirect proof is more effective:• Assume that the argument is not valid. But then there must be

a valuation that satisfies all the premises but not the conclusion:

• h p, h w | w p 1 1 0

1 0 01 0 1 0

0 contradiction

Logical entailment in PL

• All the arguments with the same logical form as a valid argument are valid:h p, h w |= w p

For variables h, p, w any elementary sentence can be substituted:He plays a piano or studies logic.If he plays a piano then he is a virtuous. If he is not a virtuous then he studies logic.

Valid argument – the same valid logical form

Logical entailment

• An argument is valid, P1,...,Pn |= Z, iffthe implicative formula is a tautology:

|= (P1 ... Pn) Z.• A proof that a formula is a tautology or that

a conclusion Z logically follows from premises can be performed:a) In the direct way – for instance by a truth-value

table (only in PL), by natural deduction, etc.b) In the indirect way: P1 ... Pn Z is a

contradiction; hence the set of premises + negated conclusion {P1, ..., Pn, Z} is contradictory, i.e., does not have a model: there is no valuation under which all the elements of the set are true.

A proof of a tautology

|= ((p q) q) pIndirect: ((p q) q) p negated f., must be a contradiction 1 1 attempt whether it can be 1

1 1 1 1 0

contradiction

There is no valuation under which the negated formula is true. Therefore, the original formula is a tautology

The most important tautologies in PL

Tautologies with one propositional variable:

|= p p

|= p p the law of excluded middle

|= (p p) the law of contradiction

|= p p the law of double negation

Algebraic laws for conjunction, disjunction and equivalence

• |= (p q) (q p) commutative laws• |= (p q) (q p)• |= (p q) (q p)

• |= [(p q) r] [p (q r)] associative laws• |= [(p q) r] [p (q r)]• |= [(p q) r] [p (q r)]

• |= [(p q) r] [(p r) (q r)] distributive laws• |= [(p q) r] [(p r) (q r)]

Laws of implication

|= p (q p) law of simplification|= (p p) q Duns Scot’s law|= (p q) (q p) law of contra-position|= (p (q r)) ((p q) r) premises joint|= (p (q r)) (q (p r)) order of premises does not

matter

|= (p q) ((q r) (p r)) hypothetic sylogism|= ((p q) (q r)) (p r) transitivity of implication|= (p (q r)) ((p q) (p r)) Frege’s law

|= (p p) p reductio ad absurdum|= ((p q) (p q)) p reductio ad absurdum

|= (p q) p , |= (p q) q|= p (p q) , |= q (p q)

Laws of transformation

|= (p q) (p q) (q p)

|= (p q) (p q) (q p)

|= (p q) (p q) (q p)

|= (p q) (p q)

|= (p q) (p q) Negation of implication

|= (p q) (p q) De Morgan law

|= (p q) (p q) De Morgan law

These laws define a method for negating

Negation of implication

Parents: If you behave well you’ll get a new ski at Christmas! (p q)

Child: I did behave well all the year and there is no ski under the Christmas tree!

p q(Did the parents fulfill their promise?)

Attorney general:If the accused man is guilty then there was an accessory in the fact

Defence lawyer:It is not true !

Question: Did the advocate (defence lawyer) help the accused man; what did he actually claim?(The man is guilty and he performed the illegal act alone!)

Negation of implicationSentence in the future tense:

If you steel it I’ll kill you! (p q)It is not true: I will steel it and yet you will not kill me. p q

OK, but: If the 3rd world war breaks out tomorrow then more than three million

people will be killed.It is not true: The 3rd world war will break out tomorrow and less than three

million people will be killed ??? Probably by negating the sentence we did not intend to claim that

(certainly) the 3rd world war will break out tomorrow:There is an unsaid modality: Necessarily, if the 3rd world war breaks

out tomorrow then more than three million people will be killed. It is not true: Possibly the 3rd world war breaks out tomorrow but at

that case less than three million people will be killed.

Handled by modal logics – not a subject of this course.

Some more arguments• Transformation from natural language may be

ambiguous:If a man has high blood pressure and breathes

with difficulties or he has a fever then he is sick.p – ”X has high blood pressure”q – ”X breathes with difficulties”r – ”X has a fever”s – ”X is sick” 1. possible analysis: [(p q) r] s

2. possible analysis: [p (q r)] s

Some more arguments

If Charles has high blood pressure and breathes with difficulties or he has a fever then he is sick.

Charles is not sick but he breathes with difficulties.

What can be deduced from these facts?We have to distinguish between first and

second reading because they are not equivalent. The conclusions will be different.

Analysis of the 1. reading

1. analysis: [(p q) r] s, s, q ???a) By means of equivalent transformations:

[(p q) r] s, s [(p q) r] (de

transposition Morgan)

(p q) r (p q), r, but q holds p, r (consequences)

Hence Charles does not have a high blood pressure and does not have a fever.

Analysis of the 2. reading

2. analysis: [p (q r)] s, s, q ??? a) reasoning with equivalent transformations:

[p (q r)] s, s [p (q r)] transposition de Morgan:

p (q r) but q is true the second disjunct cannot be true the first is true:

p (consequence)

Hence Charles does not have a high blood pressure

(we cannot conclude anything about his temperature r)

A proof of both cases

1. analysis: [(p q) r] s, s, q |= p,r2. analysis: [p (q r)] s, s, q |= p home worka) 1. case – by means of a table: home workb) Indirect: premises + negated conclusion (p r)

(p r) and we assume that every f. is true:• [(p q) r] s, s, q, p r• 1 1 0 1 1 • 0 0 • 0 0 • 0 1

p r = 0 contradiction

Summary

• Typical tasks:– Verifying a valid argument– What can be deduced from given assumptions?– Add the missing assumptions– Is a given formula a tautology, contradiction,

satisfiable?– Find models of a formula, find a model of a set of

formulas• Methods we have learnt till now for PL:

– Table method– reasoning and equivalent transformation– Indirect proof

FOPL: Interpretation, models 2004/18/23

First-order predicate logicFirst-order predicate logicWe have seen that the question

“Is a formula A true?”is reasonable only when we add

“in the interpretation I for a valuation v of free variables”. InterpretaInterpretationtion stru structurecture is an n-tuple:

I = U, R1,...,Rn, F1,...,Fm,

where F1,...,Fm are funcfunctionstions over the universe of discourse assigned to the functional symbols occurring in the formula, and

R1,...,Rn are relarelationstions over the universe of discourse assigned to the predicate symbols occurring in the formula.

How to evaluate the truth-value of a formula in an interpretation structure I, or for short in the Interpretation I?


Interpretation, evaluation of a formula

We evaluate bottom up, i.e., from the “inside out” : A. First, determine the elements of the universe denoted by terms,B. then determine the truth-values of atomic formulas, andC. finally, determine the truth-value of the (composed) formula

A. Evaluation of terms: Let v be a valuation that associates each variable x with an element of the universe: v(x) U. By evaluation e of terms induced by v we obtain an element e(x) of the universe U that is defined inductively as follows: e(x) = v(x)

e(f(t1, t2,...,tn)) = F(e(t1), e(t2),...,e(tn)),

where F is the function assigned by I to the functional symbol f.


Interpretation, evaluation of a formula

B. Evaluation of a formula1. Atomic formulas: |=I P(t1,...,tn)[v] – the formula is true in the

interpretation I for a valuation v iff

e(t1), e(t2),...,e(tn) R, where R is the relation assigned to the symbol P (we also say that R is the domain of truth of P)

2. Composed formulas:a) Propositionally composed A, A B, A B, A B, A B,

dtto Propositional Logicb) Quantified Formulas xA(x), xA(x):

|=I xA(x)[v], if for any individual i U holds |=I A[v(x/i)], where v(x/i) is a valuation identical to v up to

assigning the individual i to the variable x

|=I xA(x)[v], if for at least one individual i U holds |=I A[v(x/i)].


Quantifiers

It is obvious from the definition of quantifiers that over a finite universe of discourse U = {a1,…,an} the following equivalences hold:

x A(x) A(a1) … A(an) x A(x) A(a1) … A(an) Hence universal quantifier is a generalization of a

conjunction; existential quantifier is a generalization of a disjunction.

Therefore, the following equivalences obviously holds: x A(x) x A(x), x A(x) x A(x)

de Morgan laws

Satisfiability and validity in interpretation

Formula A is satisfiable in interpretation I, if there exists valuation v of variables that |=I A[v].

Formula A is true in interpretation I, |=I A, if for all possible valuations v holds that |=I A[v].

Model of a formula A is an interpretation I, in which A is true(that means for all valuations of free variables).

Formula A is satisfiable, if there is interpretation I, in which A is satisfied (i.e., if there is an interpretation I and valuation v such that |=I A[v].)

Formula A is a tautology (logically valid), |= A, if A is true in every interpretation (i.e., for all valuations).

Formula A is a contradiction, if there is no interpretation I, that would satisfy A, so there is no interpretation and valuation, in which A would be true: |I A[v], for any I and v.


A: x P(f(x), x) B: x P(f(x), x) C: P(f(x), x) Interpretation I: U=N, f x2, P relation > It is true that: |=I B. Formula B is in N, x2, > true. Formulas A and C are in N, >, x2 satisfied, but not true:

for e0(x) = 0, e1(x) = 1 pairs 0,0, 1,1 are not elements of >; for e2(x) = 2, e3(x) = 3, …, pairs 4,2, 9,3, … are elements of the relation >.

Formulas A, C are not in N, x2, > true: |I A[e0], |I A[e1], |I C[e0], |I C[e1],

only: |=I A[e2], |=I A[e3], |=I C[e2], |=I C[e3], …

Empty universe?

Consider an empty universe U = x P(x): is it true or not?

By the definition of quantifiers it is false, because we can’t find any individual which would satisfy P, then it is true that x P(x), so x P(x), but this is false as well – contradiction.

Or it is true, because there is no element of the universe that would not have the property P, but then x P(x) should be true as well, which is false – contradiction.

Likewise for x P(x) leads to a contradiction So we always choose a non-empty universe of

interpretation Logic “of an empty world” would not be not reasonable

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Existential quantifier + implication? There is somebody such that if he/she is a genius,

then everybody is a genius. This sentence cannot be false: |= x (G(x) xG(x)) For every interpretation I it holds:

If the truth-domain GU of the predicate G is equal to the whole universe (GU = U), then the formula is true in I, because the subformula xG(x) is true; hence G(x) x G(x), and x (G(x) xG(x)) is true in I.

If GU is a proper subset of U (GU U), then it suffices to find at least one individual a (assigned by valuation v to x) such that a is not an element of GU. Then G(a) x G(x) is true in I, because the antecedent G(a) is false. Hence x (G(x) xG(x)) is true in I.

Existential quantifier + conjunction !

Similarly x (P(x) Q(x)) is “almost” a tautology. It is true in every interpretation I such that PU U, because then |=I P(x) Q(x)[v] for v(x) PU

or QU = U, because then |=I P(x) Q(x) for all valuations

So this formula is false only in such an interpretation I where PU = U and QU U.

Therefore, sentences of a type “Some P’s are Q’s” are analyzed by x (P(x) Q(x)).

Universal quantifier + conjunction? Usually no, but implication!

Similarly x [P(x) Q(x)] is ”almost” a contradiction! The formula is false in every interpretation I such that

PU U or QU U. So the formula is true only in an interpretation I such that PU = U

a QU = U Therefore, sentences of a type

“All P’s are Q’s“ are analyzed by x [P(x) Q(x)]

It holds for all individuals x that if x is a P then x is a Q. (See the definition of the subset relation PU QU)


Formula A(x) with a free variable x: If A(x) is true in I, then |=I x A(x) If A(x) is satisfied in I, then |=I x A(x).Formulas P(x) Q(x), P(x) Q(x)

with the free variable x define the intersection and union, respectively, of truth-domains PU, QU. For every P, Q, PU, QU and an interpretation I it holds:

|=I x [P(x) Q(x)] iff PU QU

|=I x [P(x) Q(x)] iff PU QU |=I x [P(x) Q(x)] iff PU QU = U

|=I x [P(x) Q(x)] iff PU QU

Model of a set of formulas, logical entailment

A Model of the set of formulas {A1,…,An} is an interpretation I such that each of the formulas A1,...,An is true in I.

Formula B logically follows from A1, …, An, denoted A1,…,An |= B, iff B is true in every model of {A1,…,An}.

Thus for every interpretation I in which the formulas A1, …, An are true it holds that the formula B is true as well:

A1,…,An |= B: If |=I A1,…, |=I An then |=I B, for all I. Note that the “circumstances“ under which a formula is,

or is not, true (see the 1st lesson, Definition 1) are in FOL modelled by interpretations (of predicates and functional symbols by relations and functions, respectively, over the universe).

Logical entailment in FOL P(x) |= x P(x), but the formula P(x) x P(x) is obviously not a

tautology.

Therefore, A1,...,An |= Z |= (A1… An Z) holds only for closed closed formulformulasas, so-called sentencesentencess.

x P(x) P(a) is also not a tautology, and thus the rule x P(x) | P(a) is not truth-preserving;

P(a) does not logically follow form x P(x). Example of an interpretation I such that x P(x) is, and

P(a) is not true in I: U = N(atural numbers), P even numbers, a 3

Semantic verification of an argument

An argument is valid iff the conclusion is true in every model of the set of the premises.

But the set of models can be infinite! And, of course, we cannot examine an infinite

number of models; but we can verify the ‘logical form’ of the argument, and check whether the models of premises do satisfy the conclusion.

Semantic verification of an argument Example:

All monkeys (P) like bananas (Q) Judy (a) is monkey Judy likes bananas

x [P(x) Q(x)] QU

P(a) PU

Q(a) a

Relations

Propositions with unary predicates (expressing properties of individuals) were studied already in the ancient times by Aristotle.

Until quite recently Gottlob Frege, the founder of modern logic, developed the system of formal predicate logic with n-ary predicates characterizing relations between individuals, and with quantifiers.

Frege, however, used another language than the one of the current FOL.

Aristotle: (384 BC – March 7, 322 BC)

a Greek philosopher, a student of Plato and teacher of Alexander the Great.

He wrote on diverse subjects, including physics, metaphysics, poetry (including theater), biology and zoology, logic, rhetoric, politics, government, and ethics.

Along with Socrates and Plato, Aristotle was one of the most influential of the ancient Greek philosophers. They transformed Presocratic Greek philosophy into the foundations of Western philosophy as we know it.

Plato and Aristotle have founded two of the most important schools of Ancient philosophy.

37

Gottlob Frege

1848 – 1925

German mathematician, logician and philosopher, taught at the University of Jena.

Founder of modern logic.


Marie likes only winners Karel is a winner -------------------------------------- invalid Marie likes Karel

x [R(m,x) V(x)], V(k) R(m,k) ?

RU U U: { <Marie, i1>, <Marie, i2>, …, <Marie, in> …} VU U: {…i1, i2, …, Karel,…, in…}

The pair <Marie, Karel> doesn’t have to be an element of RU, it is not guaranteed by the validity of the premises.

Being a winner is only a necessary conditionnecessary condition for Marie’s liking somebody, but it is not a sufficient condition.


Marie likes only winners Karel is not a winner ------------------------------------- valid Marie does not like Karel

x [R(m,x) V(x)], V(k) R(m,k)

RU U U:{…<Marie, i1>, <Marie, i2>, <Marie, Karel>, …, <Marie, in> …}

VU U: {…i1, i2, …, Karel, Karel,…, in…}

Let the pair <Marie, Karel> be an element of RU; then by the first premise Karel has to be an element of VU, but it is not so if the second premise is true. Hence the pair <Marie, Karel> is not an element of RU.The validity of the conclusion is guaranteed by the validity of premises.


Anybody who knows Marie and Karel is sorry for Marie.x [(K(x,m) K(x,k)) S(x,m)]

Some are not sorry for Marie though they know her. x [S(x,m) K(x,m)]

|= Somebody knows Marie but not Karel. x [K(x,m) K(x,k)]

We illustrate the truth-domain of the predicates K and S, i.e., the relations KU and SU that satisfy the premises:

KU = {…, i1,m, i1,k, i2,m, i2,k,…, ,m,m,… }1. premise 2. premise

SU = {…, i1,m, ...., i2,m,…........., ,m,m,… }

Semantic verification of an argument: Semantic verification of an argument: an an indirect proofindirect proof

Anybody who knows Marie and Karel is sorry for Marie. x [(K(x,m) K(x,k)) S(x,m)]

Some are not sorry for Mariethough they know her. x [S(x,m) K(x,m)]

|= Somebody knows Marie but not Karel. x [K(x,m) K(x,k)]

Assume now that all the individuals who are paired with m in KU are also paired with k in KU:

KU = {…, i1,m, i1,k, i2,m, i2,k,…, ,m,m, ,k,k … }

SU = {…, i1,m, ...., i2,m,…........., ,m,m, ,m,m … }

contradiction

Some important tautologies of FOL|= xAx Ax/t term t is substitutable for x in A

|= Ax/t xAx De Morgan

|= x Ax x Ax|= x Ax x Ax

The laws of quantifier distribution:

|= x [A(x) B(x)] [x A(x) x B(x)]

|= x [A(x) B(x)] [x A(x) x B(x)]

|= x [A(x) B(x)] [x A(x) x B(x)]

|= x [A(x) B(x)] [x A(x) x B(x)]

|= [xA(x) xB(x)] x [A(x) B(x)]

|= x [A(x) B(x)] [x A(x) x B(x)]

Semantic proofs: Let AU, BU be truth-domains of A, Bx[A(x) B(x)] [xA(x) xB(x)]If the intersection (AU BU) = U, then AU and BU must be equal to the

whole universe U, and vice-versa.x[A(x) B(x)] [xA(x) xB(x)]If the union (AU BU) , then AU or BU must be non-empty (AU

, or BU ), and vice-versa.|= x[A(x) B(x)] [xA(x) xB(x)]If AU BU, then if AU = U then BU = U.|= x[A(x) B(x)] [xA(x) xB(x)]If AU BU, then if AU then BU .|= x[A(x) B(x)] [xA(x) xB(x)]If the intersection (AU BU) , then AU and BU must be non-empty

(AU , BU ).|= [xA(x) xB(x)] x[A(x) B(x)]If AU = U or BU = U, then the union (AU BU) = U

Some important tautologies Formula A does not contain free variable x:

|= x[A B(x)] [A xB(x)]|= x[A B(x)] [A xB(x)]|= x[B(x) A] [xB(x) A]|= x[B(x) A] [xB(x) A]|= x[A B(x)] [A xB(x)]|= x[A B(x)] [A xB(x)]|= x[A B(x)] [A xB(x)]|= x[A B(x)] [A xB(x)]

The commutative law of quantifiers.

|= xyA(x,y) yxA(x,y)|= xyA(x,y) yxA(x,y) |= xyA(x,y) yxA(x,y) but not vice-versa!

Semantic proofs: Let AU, BU be truth- domains of A, B (x not free in A

x[A B(x)] [A xB(x)] – obvious

x[A B(x)] [A xB(x)] – obvious

x [B(x) A] [x B(x) A]

x [B(x) A] x [B(x) A]: the complement BU or A is the whole universe: x B(x) A x B(x) A x B(x) A

x[B(x) A] [xB(x) A]

x [B(x) A] x [B(x) A]: the complement BU is non-empty or A: x B(x) A x B(x) A x B(x) A

Typical problems Prove the logical validity of a formula:

A formula F is true in all interpretations, which means that every interpretation is a model

|= F Prove the validity of an argument:

P1, …, Pn |= Q

for close formulas iff |= (P1 … Pn Q) formula Q is true in all the models of the set of premises

P1, …, Pn

What is entailed by the given premises? P1, …, Pn |= ?

Typical problems

Semantic solution over an infinite set of models is difficult, semantics proofs are tough.

So we are trying to find some other methods One of them is the semantic-tableau method. Analogy, generalization of the same method in

propositional logic Transformation to a disjunctive / conjunctive

normal form.

Semantic tableau in FOL

When proving a tautology by a direct proof – we use a conjunctive normal form an indirect proof – disjunctive normal form In order to apply the propositional logic method of

semantic tableau, we have to get rid of quantifiers. How to eliminate them?

To this end we use the following rules: x A(x) | A(x/t), where t is a term which is

substitutable for x in A, usually t = x (x)A(x) | A(a), where a is a new constant (not used

in the proof as yet)

Rules for quantifiers elimination

x A(x) | A(x/t), term t is substitutable for x If the truth-domain AU = U, then the individual e(t) is an element of AU

The rule is truth-preserving, OK (x)A(x) | A(a), where a is a new constant

If the truth-domain AU , the individual e(a) might not be an element of AU

The rule is not truth-preserving! x (y) B(x,y) | B(a, b), where a, b are suitable constants

Though if for every x there is a y such that the pair <x,y> is in BU, the pair <a, b> might not be an element of BU.

The rule is not truth-preserving! However, existential-quantifier elimination does not yield a contradiction: it is

possible to interpret the constants a, b so that the formula on the right-hand side is true, whenever the formula on the left-hand side is true.

For this reason we use the indirect proof (disjunctive tableau), whenever the premises contain existential quantifier(s)

Semantic tableau in FOL– disjunctive

Example. Proof of the logical validity of a formula: |= x [P(x) Q(x)] [x P(x) x Q(x)] Indirect proof (non-satisfiable of formula): x [P(x) Q(x)] x P(x) x Q(x) (order!)

x [P(x) Q(x)], P(a) Q(a), P(a), Q(a)

x [P(x) Q(x)], P(a), P(a), Q(a) x [P(x) Q(x)], Q(a), P(a), Q(a)

+ +Both branches are closed, they are contradictory. Therefore, the original

(blue) formula is tautology.

1. 2. 3.

Semantic tableau |=? x [P(x) Q(x)] [x P(x) x Q(x)] Negation:

x [P(x) Q(x)] x P(x) x Q(x)x [P(x) Q(x)], P(a), Q(b) 1.eliminaton - diff. const. !P(a) Q(a), P(b) Q(b), P(a), Q(b) 2. elimination

P(a), P(b) Q(b), P(a), Q(b) Q(a), P(b) Q(b), P(a), Q(b)

P(a), P(b), P(a), Q(b) P(a), Q(b), P(a), Q(b) Q(a), P(b), P(a), Q(b) Q(a), Q(b), P(a),

Q(b)

Formula is not logically valid, 3. branch is not closed

Tableau can lead to an infinite evaluation F: x y P(x,y) x P(x,x)

x y z ([P(x,y) P(y,z)] P(x,z)) Variable x is bound by universal quantifier We must “check all x” : a1, a2, a3, … For y we must choose always another constant:

P(a1, a2), P(a1, a1) P(a2, a3), P(a2, a2), P(a2, a1)P(a3, a4), P(a3, a3), P(a3, a2)P(a4, a5), P(a4, a4), P(a4, a3)

… The problem of logical validity is not decidable in FOL

Tableau can lead to an infinite evaluation F: x y P(x,y) x P(x,x)

x y z ([P(x,y) P(y,z)] P(x,z)) What kind of formula is F? Is it satisfiable, contradictory or logicaly valid? Try to find a model:1. U = N2. PU = relation < (less then)

1 2 3 4 5 ... satisfiable

Could the formula F have a finite model?U = {a1, a2, a3, ... ? }

To a1 there must exist an element a2, so that P(a1, a2), a2 a1

To a2 there must exist an element a3 such that P(a2, a3), a3 a2, and a3 a1

otherwise P(a1, a2) P(a2, a1), so P(a1, a1).

To a3 there must exist an element a4 such that P(a3, a4), a4 a3, and a4 a2 otherwise P(a2, a3) P(a3, a2), so P(a2, a2).And so on ad infinitum…

Argument validity- indirect proof x [P(x) Q(x)] x Q(x) |= x P(x)

x [P(x) Q(x)], x Q(x), x P(x) – contradictory?

x [P(x) Q(x)], Q(a), x P(x)

x [P(x) Q(x)], x P(x), [P(a) Q(a)], Q(a), P(a)

P(a), Q(a), P(a) Q(a), Q(a), P(a)

+ +Both branches are closed. The set of premises together with

the negated conclusion is contradictory; so the argument is valid…

Argument validity- indirect proof x [P(x) Q(x)] x Q(x) |= x P(x)

No whale is fish.

The fish exists.

Some individuals are not the whales. The set of statements: {No whale is fish, fish exists, all individuals

are whales} is contradictory.

Consistency checking

There is a barber who shaves just those who do not shave themselves

Does the barber shave himself? x y [H(y,y) H(x,y)] |= ? H(y,y) H(a,y), H(y,y) H(a,y) – eliminating H(a,a) H(a,a), H(a,a) H(a,a) – eliminating H(a,a), H(a,a) H(a,a), H(a,a), H(a,a) H(a,a)H(a,a), H(a,a) H(a,a), H(a,a) …

+The first sentence is contradictory; anything is entailed by

it. But, such a barber does not exist.

Summary – semantic tableau in FOL We use semantic tableaus for an indirect proof, i.e., transform a formula to the disjunctive normal form (branching means disjunction, comma conjunction)

There is a problem with closed formulas. We need to eliminate quantifiers.

First, eliminate existential quantifiers: replace the variable (which is not in the scope of any universal quantifier) by a new constant that is still not used.

Second, eliminate universal quantifiers: replace the universally bound variables step by step by suitable constants, until a contradiction emerges, i.e., the branch gets closed

If a variable x is bound by an existential quantifier and x is in the scope of a universal quantifier binding a variable y, we must gradually replace y by suitable constants and consequently the variable x by new, not used constants …

If the tableau eventually gets closed, the formula or a set of formulas is contradictory.

Example – semantic tableau

|= xy P(x,y) yx P(x,y) negation: xy P(x,y) yx P(x,y)

yP(a,y), xP(x,b)

x/a, y/b (for all…, hence also for a, b)

P(a,b), P(a,b)

+

Example – semantic tableau

|= [x P(x) x Q(x)] x [P(x) Q(x)] negation: [x P(x) x Q(x)] x [P(x) Q(x)]

x P(x), P(a), Q(a) x Q(x), P(a), Q(a)

xP(x),P(a),P(a),Q(a) xQ(x),Q(a),P(a),Q(a)

+ +

60

Gottlob Frege Friedrich Ludwig Gottlob Frege (b. 1848, d. 1925) was a German mathematician,

logician, and philosopher who worked at the University of Jena. Frege essentially reconceived the discipline of logic by constructing a formal

system which, in effect, constituted the first ‘predicate calculus’. In this formal system, Frege developed an analysis of quantified statements and formalized the notion of a ‘proof’ in terms that are still accepted today.

Frege then demonstrated that one could use his system to resolve theoretical mathematical statements in terms of simpler logical and mathematical notions. Bertrand Russell showed that of the axioms that Frege later added to his system, in the attempt to derive significant parts of mathematics from logic, proved to be inconsistent.

Nevertheless, his definitions (of the predecessor relation and of the concept of natural number) and methods (for deriving the axioms of number theory) constituted a significant advance. To ground his views about the relationship of logic and mathematics, Frege conceived a comprehensive philosophy of language that many philosophers still find insightful. However, his lifelong project, of showing that mathematics was reducible to logic, was not successful.

Stanford Encyclopedia of Philosophy http://plato.stanford.edu/entries/frege/

Bertrand Russell

1872-1970 British philosopher,

logician, essay-writer

Bertrand Russell

Bertrand Arthur William Russell (b.1872 - d.1970) was a British philosopher, logician, essayist, and social critic, best known for his work in mathematical logic and analytic philosophy. His most influential contributions include his defense of logicism (the view that mathematics is in some important sense reducible to logic), and his theories of definite descriptions and logical atomism. Along with G.E. Moore, Russell is generally recognized as one of the founders of analytic philosophy. Along with Kurt Gödel, he is also regularly credited with being one of the two most important logicians of the twentieth century.

Kurt Gödel (1906-Brno, 1978-Princeton)

The greatest logician of 20th century, a friend of A. Einstein, became famous by his Incompleteness Theorems of arithmetic

Russell's Paradox

Russell's paradox is the most famous of the logical or set-theoretical paradoxes. The paradox arises within naive set theory by considering the set of all sets that are not members of themselves. Such a set appears to be a member of itself if and only if it is not a member of itself, hence the paradox.

http://plato.stanford.edu/entries/russell-paradox/

Russell's Paradox

Some sets, such as the set of all teacups, are not members of themselves. Other sets, such as the set of all non-teacups, are members of themselves. Call the set of all sets that are not members of themselves "R." If R is a member of itself, then by definition it must not be a member of itself. Similarly, if R is not a member of itself, then by definition it must be a member of itself. Discovered by Bertrand Russell in 1901, the paradox has prompted much work in logic, set theory and the philosophy and foundations of mathematics.

Russell's Paradox

R – the set of all normal sets that are not members of themselves

Question: “Is R normal?” yields a contradiction. In symbols: xR (xx) – by the definition of R The question R R? yields a contradiction: RR RR, because: Answer YES – R is not normal, RR, but by the

definition R is not a member of R, i.e. RR Answer NO – R is normal, RR, but then by the

definition RR (because R is the set of all normal sets)

Russell wrote to Gottlob Frege with news of his

paradox on June 16, 1902. The paradox was of significance to Frege's logical work since, in effect, it showed that the axioms Frege was using to formalize his logic were inconsistent.

Specifically, Frege's Rule V, which states that two sets are equal if and only if their corresponding functions coincide in values for all possible arguments, requires that an expression such as f(x) be considered both a function of the argument x and a function of the argument f. In effect, it was this ambiguity that allowed Russell to construct R in such a way that it could both be and not be a member of itself.

68

Russell's Paradox

Russell's letter arrived just as the second volume of Frege's Grundgesetze der Arithmetik (The Basic Laws of Arithmetic, 1893, 1903) was in press. Immediately appreciating the difficulty the paradox posed, Frege added to the Grundgesetze a hastily composed appendix discussing Russell's discovery. In the appendix Frege observes that the consequences of Russell's paradox are not immediately clear. For example, "Is it always permissible to speak of the extension of a concept, of a class? And if not, how do we recognize the exceptional cases? Can we always infer from the extension of one concept's coinciding with that of a second, that every object which falls under the first concept also falls under the second? These are the questions," Frege notes, "raised by Mr Russell's communication." Because of these worries, Frege eventually felt forced to abandon many of his views about logic and mathematics.

Of course, Russell also was concerned about the contradiction. Upon learning that Frege agreed with him about the significance of the result, he immediately began writing an appendix for his own soon-to-be-released Principles of Mathematics. Entitled "Appendix B: The Doctrine of Types," the appendix represents Russell's first detailed attempt at providing a principled method for avoiding what was soon to become known as "Russell's paradox."

69

Russell's ParadoxThe significance of Russell's paradox can be seen once it is realized that, using

classical logic, all sentences follow from a contradiction. For example, assuming both P and ~P, any arbitrary proposition, Q, can be proved as follows: from P we obtain P Q by the rule of Addition; then from P Q and ~P we obtain Q by the rule of Disjunctive Syllogism. Because of this, and because set theory underlies all branches of mathematics, many people began to worry that, if set theory was inconsistent, no mathematical proof could be trusted completely.

Russell's paradox ultimately stems from the idea that any coherent condition may be used to determine a set. As a result, most attempts at resolving the paradox have concentrated on various ways of restricting the principles governing set existence found within naive set theory, particularly the so-called Comprehension (or Abstraction) axiom. This axiom in effect states that any propositional function, P(x), containing x as a free variable can be used to determine a set. In other words, corresponding to every propositional function, P(x), there will exist a set whose members are exactly those things, x, that have property P. It is now generally, although not universally, agreed that such an axiom must either be abandoned or modified.

Russell's own response to the paradox was his aptly named theory of types. Recognizing that self-reference lies at the heart of the paradox, Russell's basic idea is that we can avoid commitment to R (the set of all sets that are not members of themselves) by arranging all sentences (or, equivalently, all propositional functions) into a hierarchy. The lowest level of this hierarchy will consist of sentences about individuals. The next lowest level will consist of sentences about sets of individuals. The next lowest level will consist of sentences about sets of sets of individuals, and so on. It is then possible to refer to all objects for which a given condition (or predicate) holds only if they are all at the same level or of the same "type."

70

Russell’s paradox – 3 solutions

Russell's own response to the paradox was his aptly named theory of types. Recognizing that self-reference lies at the heart of the paradox, Russell's basic idea is that we can avoid commitment to R (the set of all sets that are not members of themselves) by arranging all sentences (or, equivalently, all propositional functions) into a hierarchy. The lowest level of this hierarchy will consist of sentences about individuals. The next lowest level will consist of sentences about sets of individuals. The next lowest level will consist of sentences about sets of sets of individuals, and so on. It is then possible to refer to all objects for which a given condition (or predicate) holds only if they are all at the same level or of the same "type."

This solution to Russell's paradox is motivated in large part by the so-called vicious circle principle, a principle which, in effect, states that no propositional function can be defined prior to specifying the function's range. In other words, before a function can be defined, one first has to specify exactly those objects to which the function will apply. (For example, before defining the predicate "is a prime number," one first needs to define the range of objects that this predicate might be said to satisfy, namely the set, N, of natural numbers.) From this it follows that no function's range will ever be able to include any object defined in terms of the function itself. As a result, propositional functions (along with their corresponding propositions) will end up being arranged in a hierarchy of exactly the kind Russell proposes.

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Russell’s paradox – 3 solutions

Although Russell first introduced his theory of types in his 1903 Principles of Mathematics, type theory found its mature expression five years later in his 1908 article, "Mathematical Logic as Based on the Theory of Types," and in the monumental work he co-authored with Alfred North Whitehead, Principia Mathematica (1910, 1912, 1913). Russell's type theory thus appears in two versions: the "simple theory" of 1903 and the "ramified theory" of 1908. Both versions have been criticized for being too ad hoc to eliminate the paradox successfully. In addition, even if type theory is successful in eliminating Russell's paradox, it is likely to be ineffective at resolving other, unrelated paradoxes.

Other responses to Russell's paradox have included those of David Hilbert and the formalists (whose basic idea was to allow the use of only finite, well-defined and constructible objects, together with rules of inference deemed to be absolutely certain), and of Luitzen Brouwer and the intuitionists (whose basic idea was that one cannot assert the existence of a mathematical object unless one can also indicate how to go about constructing it).

Yet a fourth response was embodied in Ernst Zermelo's 1908 axiomatization of set theory. Zermelo's axioms were designed to resolve Russell's paradox by again restricting the Comprehension axiom in a manner not dissimilar to that proposed by Russell. ZF and ZFC (i.e., ZF supplemented by the Axiom of Choice), the two axiomatizations generally used today, are modifications of Zermelo's theory developed primarily by Abraham Fraenkel.

Thank you for your attention

See you next week

Lesson 3 Validity of arguments Marie Duží. Logical entailment A formula A logically follows from a set of formulas M, denoted M |= A, iff A is true in.

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