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While a -> a and (a -> b) -> [a] -> [b] have their relation variables reduced into functions, that's not the case for (forall x. a x -> a x) and (a -> b) -> f a -> f b, so we don't end up without a equal/inequal theorem.
The problem is that unfoldFormula simply returns return . Predicate . IsMember x y $ rel for RelConsFunVar (with #3) and RelTypeConsApp. I'm not sufficiently familiar with free theorem derivation to know what they should return instead though.
The text was updated successfully, but these errors were encountered:
While
a -> a
and(a -> b) -> [a] -> [b]
have their relation variables reduced into functions, that's not the case for(forall x. a x -> a x)
and(a -> b) -> f a -> f b
, so we don't end up without a equal/inequal theorem.The problem is that
unfoldFormula
simply returnsreturn . Predicate . IsMember x y $ rel
forRelConsFunVar
(with #3) andRelTypeConsApp
. I'm not sufficiently familiar with free theorem derivation to know what they should return instead though.The text was updated successfully, but these errors were encountered: