prove $\neg p\implies (p\implies q)$ without using the deduction theorem prove $\neg p\implies (p\implies q)$ without using the deduction theorem
Axioms to be used are:
a)$A\implies(B\implies A)$
b)$(A\implies(B\implies C)\implies((A\implies B)\implies(A\implies C))$
c)$(\neg A\implies\neg B)\implies(B\implies A)$
Althogh by using the deduction thoerem the proof is very short and easy,using only the above axioms  it seams impossible
Can enybody give me at least a starting point
deduction theorem proof:
1)$\neg p$..................assumption
2)$\neg p\implies(\neg q\implies\neg p)$.............................by (a)
3)($\neg q\implies\neg p)$..........................................by 1,2 m.p
4)($\neg q\implies\neg p)\implies (p\implies q)$......................by (c)
5)$p\implies q$.......................................................by 3,4 m.p
 A: To derive $\neg p\to(p\to q)$, you will want substituting $A:=\neg p$ and $C:=p\to q$ into schema (b) to build the following for some useful $B$ .
$\qquad(\neg p\to(B\to (p\to q)))\to((\neg p\to B)\to(\neg p\to (p\to q)))$
Schema (a) lets us build $\neg p\to (\neg q\to\neg p)$, so if $B$ were $\neg q\to\neg p$ the above would be
$\qquad(\neg p\to((\neg q\to\neg p)\to (p\to q)))\to((\neg p\to (\neg q\to\neg p))\to(\neg p\to (p\to q)))$
With $\neg p\to(\neg q\to \neg p)$ available from schema (a), and $(\neg q\to\neg p)\to (p\to q)$ from schema (c) we would then just need to derive $\neg p\to((\neg q\to\neg p)\to(p\to q))$ and ...
That starts you off

a) $A\to(B\to A)$
b) $(A\to(B\to C))\to((A\to B)\to(A\to C))$
c) $(\neg A\to\neg B)\to(B\to A)$

*

*$(\neg p\to((\neg q\to\neg p)\to (p\to q)))\to((\neg p\to (\neg q\to\neg p))\to(\neg p\to (p\to q)))~$ from (b)

*$~~\neg p\to(\neg q\to\neg p)~$ from (a)

*$~~(\neg q\to \neg p)\to (p\to q)~$ from (c)

*$\text{something}$

*$\neg p\to((\neg q\to \neg p)\to (p\to q))~$ somehow

*Continue.

