fitch-proofs

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 ∨

问题 I'm trying to construct a formal proof for ((p ⇒ q) ⇒ p) ⇒ p. in Fitch. I know this is true, but how do I prove it? I can only use And Intro, And Elim, Or Inro, Or Elim, Neg Intro, Neg Elim, Impl Intro, Impl Elim, Biconditional Intro, and Biconditional Elim. 回答1: The following proof uses Klement's Fitch-style proof checker. Description of the symbols and the rules are in forallx . Links to both are below. A slightly different version is on Philosophy Stack Exchange: https://philosophy