It seems we have just one rule to eliminate variables: substitution. For example, given A=BC and BCD=E, we can eliminate BC by substituting A for BC in BCD=E. Thus, we must have equation !A=X to get to B!A=!A, and to get to !A=X we must have !A=Y, and so on.
So it seems impossible in given axiomatic system to derive B!A=!A from !B=AD. Am I missing something?
EDIT: Here I take axioms in 1.12 as a basis for proposition calculus and I don’t use any interpretation of them.
It seems we have just one rule to eliminate variables: substitution. For example, given A=BC and BCD=E, we can eliminate BC by substituting A for BC in BCD=E. Thus, we must have equation !A=X to get to B!A=!A, and to get to !A=X we must have !A=Y, and so on.
So it seems impossible in given axiomatic system to derive B!A=!A from !B=AD. Am I missing something?
EDIT: Here I take axioms in 1.12 as a basis for proposition calculus and I don’t use any interpretation of them.
Perhaps what is missing is these rules:
AT = A (1)
AF = F (2)
A + T = T (3)
A + F = A (4)
Which can be derived from the given axioms, apparently. I’m not sure if some necessary axioms were omitted.
Using some of these, here’s one way to derive B!A=!A from !B=AD:
!B = AD
!B + A = AD + A
!B + A = AD + AT (1)
!B + A = A(D + T) (Distributivity)
!B + A = AT (3)
!B + A = A (1)
!!B!A = !A (Duality)
B!A = !A