User:Pluke/philosophy/Logic

Truth = Property of Sentences

Validity = Property of Arguments

Logically Consistent = all sentences can be true at the same time

Logically InConsistent = not all sentences can be true at the same time

A B - C This is valid only if {A,B,¬C} is inconsistent

De Morgan's Law ¬(A V B) = ¬A ^ ¬B

IF THEN (implies) A->B =  ¬A v B

A B | R F F  T F T   T T F   F T T   T

Law of the excluded middle: A V ¬A = 1 Principle of non-Contradiction ¬(A ^ ¬A) = T

(A -> ¬A) -> ¬A = TRUE!

Modus Ponems A->B   (A V B) A       ¬A -      - B       B

IFF <-> == A->B ^ B->A

Proofs
¬(A ^ B)    (A ^ B)     (A v B)     A->B     ¬(A->B)      ¬(A v B)        / \          A          / \       / \         A            ¬A ¬A ¬B         B         A   B    ¬A   B       ¬B            ¬B (demorgans!)                                             (demorgans!)

A <--> B     ¬(A <--> B)       /  \            /  \ A  ¬A         ¬A   A     B   ¬B          B  ¬B


 * 1) Find your statement
 * 2) Take the negative of it.
 * 3) If all negative routes close then it is valid

A) All A are B - universal positive I) Some A are B - particular positive (might be all) E) No A are be - universal negative O) Some A are not B - particular negative

A) ∀x(Bx -> Cx) I)  ∃x(Bx ^ Cx) E) ¬∃x(Bx ^ Cx)  ==   ∀x(Bx -> ¬Cx) O) ¬∀x(Bx -> Cx)  ==   ∃x(Bx ^ ¬Cx)

∀x φ(x)     a replaces all (free) occurrences of of x in φ φ(a)      the name has to occur already in a branch of the tableux

∃x φ(x)     a replaces all (free) occurrences of of x in φ φ(a)     a does not occur already on the branch, it has to be new

¬∀x φ(x)   converts to above ∃x ¬φ(x)

¬∃x φ(x)   converts to above ∀x ¬φ(x)

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