Thanks, discriminate
does the job for me. My updated solution:
Theorem andb_true_elim2'' : forall b c : bool,
andb b c = true -> c = true.
Proof.
intros [].
- simpl andb.
intros c H.
apply H.
- discriminate.
Qed.
As for easy
and now
, they are too much magic for me right now. So, I'll leave them for future usage.
Jiahong Lee has marked this topic as resolved.
Last updated: Feb 08 2023 at 23:03 UTC