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: Dec 21 2024 at 16:20 UTC