Fitch proof for $(p \implies (q \implies r)) \implies ((p \implies q) \implies (p \implies r))$ with no premises I'm having trouble solving this problem using the Fitch system. As I understand Fitch, if the goal has the form $(φ \implies ψ)$, it is often good to assume $φ$ and prove $ψ$ and then use Implication Introduction to derive the goal. Since there are no premises, to prove $(p \implies (q \implies r)) \implies ((p \implies q) \implies (p \implies r))$ with the Fitch system, I'll need to assume the antecedent $(p \implies (q \implies r))$ and use Implication introduction to derive the consequent $((p \implies q) \implies (p \implies r))$. I'm stuck though as using Stanford's Fitch system, assuming $(p \implies (q \implies r))$ results in $p(q)$ and assuming $p \implies q \implies r$ results in $p$. How should I think about this problem so I can resolve the proof?

1) $p \to (q \to r)$ --- assumed [a]

2) $p \to q$ --- assumed [b]

3) $p$ --- assumed [c]

4) $q$ --- from 3) and 2) by $\to$-elim

5) $q \to r$ --- from 3) and 1) by $\to$-elim

6) $r$ --- from 4) and 5) by $\to$-elmi

7) $p \to r$ --- from 3) and 6) by $\to$-intro, discharging [b]

> 8) $(p \to q) \to (p \to r)$ --- from 2) and 7) by $\to$-intro, discharging [b]
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>> 9) $(p \to (q \to r)) \to [(p \to q) \to (p \to r)]$ --- from 1) and 8) by $\to$-intro, discharging [a].

As you can see, we have to assume all the antecedents of the conditionals involved.