Use the Fitch System to prove (¬p ⇒ q) ⇒ ((¬p ⇒ ¬q) ⇒ p).

Question

1.  p                                 Assumption
2.  ~p => ~q                          Assumption
3.  p                                 Reiteration: 1
4. (~p => ~q) => p                    Implication Introduction: 2, 3
5.  ~p => q                           Assumption
6. (~p => ~q) => p                    Reiteration: 4
7. (~p => q) => (~p => ~q) => p       Implication Introduction: 5, 6

It does not work in this way. So, I suppose p is not the right assumption...

Could you please help me to define the initial assumption to proceed?

Answer 1

We seek to prove: $\vdash (\neg p\to q)\to((\neg p\to\neg q)\to p)$

A method to prove $\phi\to (\psi\to \chi)$, would be to assume $\phi$ and $\psi$ then make an argument to conclude $\chi$, then discharge the assumptions with conditional introduction.

So then, can you make an argument for $p$ made under an assumption of $\neg p\to q$ and $\neg p \to\neg q$ ?

$$\begin{array}{r|l:l} 1. & \quad \neg p\to q & \text{Assumption}\\ 2. & \qquad \neg p\to\neg q & \text{Assumption}\\ \vdots & \qquad\quad\vdots & \vdots \\ n. & \qquad p & \\ n+1. & \quad (\neg p\to\neg q)\to p&2,n,\text{Conditional Introduction}\\ n+2. & (\neg p\to q)\to((\neg p\to\neg q)\to p)&1,n+1,\text{Conditional Introduction}\\ \end{array}$$

Answer 2

Hint

Assume $\lnot p$ and derive both $q$ and $\lnot q$.

Having reached a contradiction, we can use it to derive $\lnot \lnot p$ by $\lnot$-intro, discharging the assumption, and conclude by Double Negation.