Advertisement
Not a member of Pastebin yet?
Sign Up,
it unlocks many cool features!
- theorem ex1 : (A ⇒ C) ⇒ (B ⇒ C) ⇒ A ∨ B ⇒ C.
- apply rule (prove ((A ⇒ C) ⇒ (B ⇒ C) ⇒ A ∨ B ⇒ C));
- apply rule (⇒#i [H1] ((B⇒ C) ⇒ A ∨ B ⇒ C));
- apply rule (⇒#i [H2] (A ∨ B ⇒ C));
- apply rule (⇒#i [H3] (C));
- apply rule (∨#e (A ∨ B) [H01] (C) [H02] (C));
- [apply rule (discharge [H3]);
- |apply rule (⇒#e (A ⇒ C) (A));
- [apply rule (discharge [H1]);
- |apply rule (discharge [H01]);
- ]
- |apply rule (⇒#e (B ⇒ C) (B));
- [apply rule (discharge [H2]);
- |apply rule (discharge [H02]);
- ]
- ]
- qed.
Advertisement
Add Comment
Please, Sign In to add comment
Advertisement