Closed strub closed 1 month ago
If not, unfolding an high-order operator will produce a formula that is not convertible to the original one.
fix #590
I do not recall which context this one was from, but if it fixes something, I'm happy for it to be merged. (It does not sort out the issues I can't minimise from the nsl proof, if it was meant to...)
nsl
If not, unfolding an high-order operator will produce a formula that is not convertible to the original one.
fix #590