Semantics of FOPL

1
Lesson 6

• Semantics of FOPL
• Interpretation, models, semantic tableau

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 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, and
• R1,...,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?

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 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.

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 the domain of truth of P)
• 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 ? U holds
  I Av(x/i),
• where v(x/i) is a valuation identical to v up
  toassigning the individual i to the variable x
• I ?xA(x)v, if for at least one individual i ?
  U holds I Av(x/i).

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

6
Satisfiability and validness in interpretation

• Formula A is satisfiable in interpretation I, if
  there exists valuation v of variables that I
  Av.
• Formula A is true in interpretation I, I A, if
  for all possible valuations v holds that I
  Av.
• 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 Av.)
• 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 Av, 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 UN, f ? x2, P ? relation gt
• It is true that I B. Formula B is in ?N, x2,
  gt? true.
• Formulas A and C are in ?N, gt, x2? satisfied, but
  not true
• for e0(x) 0, e1(x) 1 these ?0,0?, ?1,1? are
  not the elements of gt, but for e2(x) 2, e3(x)
  3, the couples are ?4,2?, ?9,3?, the elements
  of relation gt.
• Formulas A, C are not in ?N, x2, gt? true
  ?I Ae0, ?I Ae1, ?I Ce0, ?I Ce1,
• only I Ae2, I Ae3, I Ce2, I
  Ce3,

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 cant 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.

10
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 Ps are Qs
• 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 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 Ps are Qs
• 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)

12
Satisfiability and validness in interpretation

• 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 ? ?

13
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).

14
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

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 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.

18
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.

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 ltMarie, i1gt, ltMarie, i2gt, ,
  ltMarie, ingt
• VU ? U i1, i2, , Karel,, in
• The pair ltMarie, Karelgt doesnt 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
  Maries 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
• ltMarie, i1gt, ltMarie, i2gt, ltMarie, Karelgt, ,
  ltMarie, ingt
• VU ? U i1, i2, , Karel, Karel,, in
• Let the pair ltMarie, Karelgt 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 ltMarie, Karelgt 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 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?,
• 1. premise 2. premise
• SU , ?i1,m?, ...., ?i2,m?,.........,
  ??,m?,

23
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?, ??,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

• ?xA(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.
• ?xA(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.
• ?xA(x) ? B(x) ? ?xA(x) ? ?xB(x)
• If AU ? BU, then if AU U then BU U.
• ?xA(x) ? B(x) ? ?xA(x) ? ?xB(x)
• If AU ? BU, then if AU ? ? then BU ? ?.
• ?xA(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) ? ?xA(x) ? B(x)
• If AU U or BU U, then the union (AU ? BU) U

26
Some important tautologies

• Formula A does not contain free variable x
• ?xA ? B(x) ? A ? ?xB(x)
• ?xA ? B(x) ? A ? ?xB(x)
• ?xB(x) ? A ? ?xB(x) ? A
• ?xB(x) ? A ? ?xB(x) ? A
• ?xA ? B(x) ? A ? ?xB(x)
• ?xA ? B(x) ? A ? ?xB(x)
• ?xA ? B(x) ? A ? ?xB(x)
• ?xA ? 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 be truth- domains of
A, B, x is not free in A

• ?xA ? B(x) ? A ? ?xB(x) obvious
• ?xA ? 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
• ?xB(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

28
Semantic tableau in predicate logic

• Proofs of logical validity and argument 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
  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 ?

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 of the same method in
  propositional logic
• Transformation to a disjunctive / conjunctive
  normal form.

31
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)

32
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 ltx,ygt is in BU, the pair lta, bgt 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
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.

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)
• Formula is 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))
• 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

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 logicaly valid?
• Try to find a model
• U N
• PU relation lt (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

37
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

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 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.

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) ? ?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/

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 Theorems of 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 definition of R
• The question R ? 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.

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 Paradox

• The 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
Russells 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
Russells 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.