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FOPL Interpretation and Evaluation - Lesson 6

Learn about the semantics of First-Order Predicate Logic (FOPL) interpretation and models, and how to evaluate the truth-value of formulas in an interpretation. Covers semantics tableaus and evaluation rules.

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FOPL Interpretation and Evaluation - Lesson 6

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  1. Lesson 6 Semantics of FOPL Interpretation, models, semantic tableau FOPL: Interpretation, models

  2. 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 vof free variables”. • Interpretation structureis ann-tuple: I= U, R1,...,Rn, F1,...,Fm, where F1,...,Fmarefunctionsover the universe of discourse assigned to the functional symbols occurring in the formula, and R1,...,Rnare relationsover the universe of discourse assigned to thepredicate 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? FOPL: Interpretation, models

  3. Interpretation, evaluation of a formula • We evaluate bottom up, i.e., from the “inside out” : • First, determine the elements of the universe denoted by terms, • then determine the truth-values of atomic formulas, and • finally, determine the truth-value of the (composed) formula • Evaluation of terms: Let v be avaluationthat 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)), whereFis the function assigned by I to the functional symbol f. FOPL: Interpretation, models

  4. Interpretation, evaluation of a formula • Evaluation of a formula • 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 thedomain of truthofP) • Composed formulas: • Propositionally composedA, A  B, A  B,A  B, A  B, dtto Propositional Logic • Quantified Formulas xA(x), xA(x): |=IxA(x)[v], if for any individual i  Uholds |=IA[v(x/i)], wherev(x/i) is a valuation identical to vup toassigning the individual i to the variable x |=IxA(x)[v], if for at least one individual i  Uholds |=IA[v(x/i)]. FOPL: Interpretation, models

  5. 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 FOPL: Interpretation, models

  6. Satisfiability and validness in interpretation • Formula A is satisfiable in interpretationI, if thereexistsvaluation vof variables that |=I A[v]. • Formula A istrue in interpretation I, |=I A, ifforallpossible valuationsvholds that |=I A[v]. • Model of formula Ais interpretation I, in which is A true(that meansfor all valuations of free variables). • Formula A is satisfiable, if there is interpretation I, in which Aissatisfied(i.e., if there is an interpretation I and valuationvsuch that |=I A[v].) • Formula A isa tautology (logically valid), |= A, if A is true in every interpretation (i.e., for all valuations). • Formula A isa contradiction, if there is nointerpretation I, that would satisfy A, sothere is no interpretation and valuation, in which A would be true: |I A[v], for any I and v.

  7. Satisfiability and validness in interpretation 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. • FormulasA and C are inN, >, x2satisfied, but not true: • fore0(x) = 0, e1(x) = 1 these0,0, 1,1 are not the elements of>, butfore2(x) = 2, e3(x) = 3, … the couples are4,2, 9,3, …the elements of relation >. • Formulas A, Care 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], …

  8. 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 xP(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 GUof 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 GUis a proper subset of U (GU U), then it suffices to find at least one individual a(assigned by valuation vtox) 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.

  10. Existential quantifier +conjunction ! • Similarly x (P(x)  Q(x))is“almost”a tautology. It is true in every interpretationI 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’sare Q’s” • are analyzed by x (P(x)  Q(x)).

  11. 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 thatPU 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’sare Q’s“ are analyzed byx [P(x)  Q(x)] • It holds for all individuals x thatif x is a P then x is a Q. (See the definition of the subset relation PU QU)

  12. Satisfiability and validness in interpretation Formula A(x) witha free variablex: • 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 definethe intersection andunion, respectively, of truth-domains PU, QU. For every P, Q, PU, QU andan 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  

  13. Model of a set of formulas, logical entailment • A Model of the set of formulas{A1,…,An} isan interpretation I such that each of the formulas A1,...,An is true in I. • Formula B logicallyfollows from A1, …, An, denotedA1,…,An|= B, iff B is true in every model of {A1,…,An}. • Thus for every interpretation I in which the formulas A1, …, Anare true it holds that the formula B is true as well: • A1,…,An|= B: If |=I A1,…, |=I Anthen |=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 andfunctional symbolsby relations andfunctions, respectively, over the universe).

  14. Logical entailment inFOPL • 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

  15. 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.

  16. 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 --------------------a Q(a)

  17. Relations • Propositions withunary 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 logicwithn-ary predicatescharacterizing relations between individuals, and with quantifiers. • Frege, however, used another language than the one of the current FOPL.

  18. Aristotle: (384 BC – March 7, 322 BC) • a Greekphilosopher, 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 PresocraticGreek 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 andphilosopher, taught at the University of Jena. Founder of modern logic.

  20. Semantic verification of an argument • 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 anecessary condition for Marie’s liking somebody, but it is not a sufficient condition.

  21. 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.

  22. Semantic verification of an argument Anybody who knows Marie and Karelis 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 andS, i.e., the relationsKU andSUthat satisfy the premises: • KU={…, i1,m,  i1,k, i2,m, i2,k,…, ,m,… } 1. premise 2. premise • SU={…, i1,m, ...., i2,m,…........., ,m,… }

  23. Semantic verification of an argument: an indirect proof Anybody who knows Marie and Karelis 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, ,k … } SU={…, i1,m, ...., i2,m,…........., ,m, ,m … } contradiction

  24. Some important tautologies |= xAx Ax/t term t is substitutable for x in A |= Ax/txAx • De Morgan |= x Axx Ax |= x Axx Ax • 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)]

  25. 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 AUand BUmust be equal to the whole universe U, andvice-versa. x[A(x)  B(x)]  [xA(x) xB(x)] If the union (AU BU)  , then AUor BU must be non-empty (AU , or BU ),andvice-versa. |= x[A(x)  B(x)]  [xA(x) xB(x)] If AU BU,thenif 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 AUand BU must be non-empty(AU , BU ). |= [xA(x) xB(x)] x[A(x)  B(x)] IfAU = U or BU = U, then the union (AU BU) = U

  26. Some important tautologies • FormulaA 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. |= xyA(x,y) yxA(x,y) |= xyA(x,y) yxA(x,y) |= xyA(x,y) yxA(x,y)but not vice-versa!

  27. Semantic proofs: Let AU, BU betruth- 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]  [xB(x)  A] x[B(x)  A]x[B(x)  A]: the complement BU or A is the whole universe: xB(x)  A xB(x)  AxB(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 xB(x) A

  28. Semantic tableau in predicate logic Proofs of logical validity andargument validity in 1st-order predicate logic

  29. Typical problems • Prove the logical validity of a formula: • A formula F is true in all interpretations,which meansthat every interpretation is a model • |= F • Prove the validity of an argument: • P1, …, Pn |= Q • forcloseformulas 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 |= ?

  30. 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 ofthe same method in propositional logic • Transformation toa disjunctive / conjunctive normal form.

  31. Semantic tableau inFOPL • 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 substitutablefor x in A, usually t = x • (x)A(x) |A(a),where a is a new constant(not used in the proof as yet)

  32. Rules for quantifiers elimination • x A(x) | A(x/t),term t issubstitutableforx • Ifthe truth-domainAU= 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-domainAU  , 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 everyxthere is ay such that the pair<x,y> is inBU, 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)

  33. 2. 3. 1. Semantic tableau inFOPL– 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.

  34. 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) Formulais not logically valid, 3. branch is not closed

  35. 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)) • Variablex 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

  36. 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 logicalyvalid? • Try to find a model: • U = N • PU = relation < (less then) 1 2 3 4 5 ... satisfiable Could the formula F have a finitemodel? U = {a1,a2, a3, ... ?} To a1 there must exist an element a2,sothat P(a1, a2), a2 a1 To a2 there must exist an element a3such 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 a4such 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…

  37. Argument validity- indirect proof • x [P(x) Q(x)] x Q(x) |= x P(x) • x [P(x) Q(x)], x Q(x), xP(x)– contradictory? x [P(x) Q(x)], Q(a), xP(x) x [P(x) Q(x)], xP(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. Theset of premises together with the negated conclusion is contradictory; so the argument is valid…

  38. 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.

  39. 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.

  40. Summary – semantic tableauinFOPL • We use semantic tableaus for an indirect proof, i.e., transform a formula to the disjunctivenormal 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.

  41. Example – semantic tableau • |= xy P(x,y)yx P(x,y) • negation:xy P(x,y)yxP(x,y) yP(a,y), xP(x,b) x/a, y/b (for all…, hence also for a, b) P(a,b), P(a,b) +

  42. Example – semantic tableau • |= [x P(x) xQ(x)] x[P(x) Q(x)] • negation:[x P(x) xQ(x)]x[P(x)  Q(x)] x P(x), P(a), Q(a)xQ(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 showedthat 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/

  44. Bertrand Russell • 1872-1970 • British philosopher, logician, essay-writer

  45. 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.

  46. Kurt Gödel (1906-Brno, 1978-Princeton) The greatest logician of 20th century, a friend of A. Einstein, became famous by his Incompleteness Theoremsof arithmetic

  47. 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/

  48. 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.

  49. 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: xR (xx) – by the definitionof R • The questionR  R?yields a contradiction: • RR RR, because: • Answer YES – R is not normal, RR, but by the definition R is not a member of R, i.e. RR • Answer NO – R is normal, RR, but then by the definition RR(because R is the set of all normal sets)

  50. 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.

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