How to prove this tautology using equivalences? I am trying to prove that the following is a tautology:
$(A \implies (B \implies C)) \implies ((A \implies (C \implies D)) \implies (A \implies (B \implies D)))$
To make progress, I thought I'd eliminate all the arrows. After that, and some de Morgan, I've arrived at:
$(A \land B \land ¬C) \lor (A \land ¬C \land ¬D) \lor (¬A \lor B \lor D) $
At this point, I don't know how to carry on, though. I feel like I'm missing some rule -- I get stuck in trying to expand this and don't get anywhere.
I'd be really grateful for help / hints!
EDIT:
Thank you Henning Makholm and Mauro ALLEGRANZA for spotting mistakes in my reformulations. The rewritten form should read:
$(A \land B \land ¬C) \lor (A \land C \land ¬D) \lor (¬A \lor ¬B \lor D) $

There must be something wrong with your rewritings -- what you have arrived at is false when all of the propositional variables are true (as well as when they're all false), whereas the original formula is true in that case.
