How To Proof of "Axioms" of Propositional Logic: Synopsis.

[Talanum1]

From (A)->-(B) (B)->-(A) we can infer "False" as follows:

1 (A)->-(B) (B)->-(A) Premise
2 )->- (A)->-(B) )->- (B)->-(A) 1, AtI
3 )->-(A) []->-(B)->-( )->-(B) []->-(A)->-( 2, A:ADx2
4 )->-(A)->-[] (B) ->-( )->-(B) []->-(A)->-( 3, Rearrange
5 )->-(A)->-[] []->-(A)->-( 4, A:AA
6 (A)->-( )->-(A) 5, A:ASS, A:SDx2
7 (_) 6, A:AA

and the empty structure is always "False".

 

[Talanum1]

Even though "(A) ->-( (A)->-(B)" seems to be writable as "(A)->-( (x)--((A)->-(B))", this is not true since then we can derive "A" from "(A)->-(B)", which does not follow.

 

[Talanum1]

We need the following axioms to prove Modus Tollens:

A;AN (B)->-( ).->-(~B) <-> (~(_))

A:NA (B)->-( (~(_)) <-> (~B)

We prove Modus Tollens (T:MT):

Line no. Statement Reason
1 (A)->-(B) (~B) Premise
2 )-.>-((A)->-(B)) )->-(~B) 1, A:AtI
3 )->-(A)->-[] (B)->-( )-.>-(~B) 2, A:AD
4 )->-(A)->-{] (~(_)) 3, A:AN
5 []->-(A)->-( (~(_)) 4, A:ASS
6 (A)->-( (~(_)) 5, A:SD
7 (~A) 6, A:NA