Download - 10/25/2015FOPL: Interpretation, models 1 Lesson 6 Semantics of FOPL Interpretation, models, semantic tableau

04/20/23 FOPL: Interpretation, models 1

Lesson 6

Semantics of FOPL

Interpretation, models, semantic tableau

FOPL: Interpretation, models 204/20/23

Truth of a formula, interpretation, evaluation

We have seen (in Lesson 4) 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”. Interpretation structure is an n-tuple:

I = U, R1,...,Rn, F1,...,Fm, where F1,...,Fm are functions over the universe of discourse assigned

to the functional symbols occurring in the formula, andR1,...,Rn are relations 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 formulaB. Evaluation of a formula

1. 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 the universal quantifier is a generalization of a

conjunction; existential quantifier is a generalization of a disjunction.

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

de Morgan laws

Satisfiability and validness 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 formula A is interpretation I, in which is A 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 these 0,0, 1,1 are not the

elements of >, but for e2(x) = 2, e3(x) = 3, … the couples are 4,2, 9,3, … the elements of 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 universum? 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

9

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 FOPL modelled by interpretations (of predicates and functional symbols by relations and functions, respectively, over the universe).

Logical entailment in FOPL

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 in FOPL only for closed formulas, so-called sentences.

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.


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

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

P(a) PU

-------------------- a

Q(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 FOPL.

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.

19

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 elements of RU, it is not guaranteed by the validity of the premises.

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

Semantic verification of an argument 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 Mariethough 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 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: an indirect 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

Semantic proofs: Let AU, BU be truth-domains of A, B

x[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

Semantic proofs: Let AU, BU be truth- domains of A, B, x is 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

Semantic tableau in predicate logic

Proofs of logical validity and argument validity in 1st-order predicate logic

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 FOPL 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 FOPL– 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

FOPL

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, but

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 FOPL 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)

+ +

43

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.

51

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."

52

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."

53

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.

54

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.

Download - 10/25/2015FOPL: Interpretation, models 1 Lesson 6 Semantics of FOPL Interpretation, models, semantic tableau

Top Related