Open salmans opened 9 years ago
On the following theory
exists x y. f(x) = y f(x) = y => g(y) = x g(x) = y => f(y) = x
exists x y. f(x) = y
f(x) = y => g(y) = x
g(x) = y => f(y) = x
@salmans poking this as well, is this still an issue?
On the following theory
exists x y. f(x) = y
f(x) = y => g(y) = x
g(x) = y => f(y) = x