L esson 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 every model of the set M.
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Lesson 3
Validity of arguments
Marie Duží
h, p, w |h p, hw| w p
1 1 1 1 11conclusion
1 1 0 1 01
1 0 1 1 11is true in all
1 0 0 1 00
0 1 1 1 11the four models
0 1 0 1 11ofpremises
0 0 1 0 11
0 0 0 0 10
h p, h w| w p
1 10
1 0 0
1 01 0
0
contradiction
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
|= (P1 ... Pn) Z.
|= ((p q) q) p
Indirect:
((p q) q) pnegated f., must be a contradiction
1 1attempt 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
Tautologies with one propositional variable:
|= p p
|= p pthe law of excluded middle
|= (p p)the law of contradiction
|= p pthe law of double negation
|= p (q p)law of simplification
|= (p p) qDuns 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) preductio ad absurdum
|= ((p q) (p q)) preductio ad absurdum
|= (p q) p , |= (p q) q
|= p (p q) , |= q (p q)
|= (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
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!)
Sentence 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.
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
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.
[(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.
[p (q r)] s,s [p (q r)]
transposition de Morgan:
p (q r) butqis true the second disjunct cannot be true the first is true:
p (consequence)
HenceCharles does not have a high blood pressure
(we cannot conclude anything about his temperature r)
1. analysis: [(p q) r] s, s, q |= p,r
2. analysis: [p (q r)] s, s, q |= phome work
p r = 0contradiction
We have seen 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 symbolsoccurring in the formula.
How to evaluate the truth-value of a formula in an interpretation structure I, or for short in the Interpretation I?
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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.
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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 composedA, A B, A B,A B, A B, dtto Propositional Logic
Quantified Formulas xA(x), xA(x):
|=IxA(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
|=IxA(x)[v], if for at least oneindividual i Uholds |=IA[v(x/i)].
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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) xA(x), x A(x) xA(x)
de Morgan laws
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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 a formula Ais an interpretation I, in which A is 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.
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 inN, >, x2 satisfied, but not true:
fore0(x) = 0, e1(x) = 1 pairs0,0, 1,1 are not elements of>;fore2(x) = 2, e3(x) = 3, …, pairs4,2, 9,3, …are elements of the relation >.
Formulas A, Care not inN, 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], …
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), sox 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 forxP(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
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.
27
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)).
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 byx [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)
Formula A(x) witha free variablex:
IfA(x) is true in I, then |=Ix A(x)
If A(x) is satisfied in I, then|=Ix 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
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 FOL modelled by interpretations (of predicates andfunctional symbolsby relations andfunctions, respectively, over the universe).
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 formulas, so-called sentences.
x P(x) P(a) is also not a tautology, and thus the rulex 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
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
Q(a) a
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 FOL.
1848 – 1925
German mathematician, logician andphilosopher, taught at the University of Jena.
Founder of modern logic.
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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 anecessary 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 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,… }
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
|= xAx Ax/tterm t is substitutable for x in A
|= Ax/txAx
De Morgan
|= x Axx Ax
|= x Axx 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)]
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
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.
|= xyA(x,y) yxA(x,y)
|= xyA(x,y) yxA(x,y)
|= xyA(x,y) yxA(x,y)but not vice-versa!
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: xB(x) A xB(x) AxB(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: xB(x) A x B(x) A xB(x) A
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 |= ?
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.
When proving a tautology by
a direct proof– we use a conjunctivenormal 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)
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)
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.
2.
3.
1.
|=? x [P(x) Q(x)] [x P(x) x Q(x)]
Negation:
x [P(x) Q(x)] xP(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
F: x y P(x,y) x P(x,x)
x yz([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 FOL
F: x y P(x,y) x P(x,x)
x yz([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…
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…
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.
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.
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.
|= xy P(x,y)yx P(x,y)
negation:xy P(x,y)yxP(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)
+
|= [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)
++
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/
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1872-1970
British philosopher, logician, essay-writer
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.
The greatest logician of 20th century, a friend of A. Einstein, became famous by his Incompleteness Theoremsof arithmetic
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/
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.
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 definitionof R
The questionR 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.
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."
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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 PQ by the rule of Addition; then from PQ 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."
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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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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.
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