问题
FYI, the logic program I use cannot do contradiction introductions. This point is most likely irrelevant, for I highly doubt I would need to use any form of contradiction for this proof.
In my attempt to solve this, I started off with assuming (p ⇒ q) ⇒ p)
Is this correct?
If so, what next? Forgive me if the solution seems so obvious.
回答1:
(p ⇒ q) ⇒ p
((p ⇒ q) ⇒ p) ∨ (p ⇒ p) ; (X ⇒ X) and Or introduction
((p ⇒ q) ∨ p) ⇒ p ; (X ⇒ Z) ∨ (Y ⇒ Z) |- (X ∨ Y ⇒ Z)
((¬p ∨ q) ∨ p) ⇒ p ; (p ⇒ q) ⇔ (¬p ∨ q)
((¬p ∨ p) ∨ q) ⇒ p ; (X ∨ Y) ∨ Z |- (X ∨ Z) ∨ Y
(true ∨ q) ⇒ p ; (¬X ∨ X) ⇔ true
true ⇒ p ; (true ∨ X) ⇔ true
p ; Implication elimination
((p ⇒ q) ⇒ p) ⇒ p ; Implication introduction
来源：https://stackoverflow.com/questions/42286985/how-would-one-prove-p-%e2%87%92-q-%e2%87%92-p-%e2%87%92-p-using-the-fitch-system