論理学ノート2

まずは定理の証明から

Theorem LetΣbe a set of wff’s， α，βalso wffs.
Then Σ，α TE(以降Tautologically Equiv.をこう表記）β iff ΣTEα->β

"->" Σ，α TEβmeans that whenever we have a truth assignment U， whatever U’(sigma)=T (for all sigma inside Σ) and U’(α)=T， then U’(β)=T also.
Now show that whenever U’(sigma)=T for all sigma in Σ， then U’(α->β)=T also.
If α is false， done. We only need to show if α is true， βalso true.
U’(α->β)=T means that whenever U’(α)=T， then U’(β) =T also. This is what we given by Σ，α TEβ

"β　Prove by Contradiction.
Assume Σ，α NTEβ
Σ，αNTEβ means there is a truth assignment s.t U’(sigma) for all sigma in Σ， and U’(α)=T but U’(β)=F.
But this means for this U， U’(α->β)=F whereas U’(sigma)=T for all sigma in Σ
Contradiction. Q.E.D

Example
a->b，b->c TE a->c is reasonable， but
b->(a->c)TE b->c is false.
Why? Make a truth assignment where U’(a)=T and U’(c)=F
Set a s.t U’(a)=F， and you get U’(b->(a->c))=T， but U’(b->c)=F
for example， a=c=A and notA， b=A or notA

うー、打つだけで疲れた。追記予定