Derive each conclusion These are short but they represent so

Derive each conclusion. These are short, but they represent some patterns of derivation that come up frequently. (Use implicational rules and equivalence rules.)

Solution

6)

C and (~A)

Therefore ~(C->~(~A)) as p and q = ~(p~->~(q))

Thus, ~(C->A) as ~(~p)=p

7) C->A, C v B B-> A

(C->A) and (B-> A) is true as both C->A and B->A are given to be true

But (pr)(qr)(pq)r

So (C->A) and (B->A) = (C v B)->A

C v B is true

Hence, T->A

Thus A (as T->p = p)