Is it possible to see how metis manages to finish a proof?
My current proof ends with:
apply (auto simp add: fun_eq_iff morph_comp_def split: if_splits)
apply metis
by metis
And I would like to drop the two metis calls.
There are metis_trace and metis_verbose, but I don't know how to read the output,
e.g. using [[metis_trace]] by metis.
Robert Soeldner has marked this topic as resolved.
Last updated: Nov 05 2025 at 20:25 UTC