#745 Anon Comment 745 A function a->b? The way I am reading what has been said is that it would give b^a, rather than a^b because a -> 1 should have 1^a == 1 possible states and 1->b should have b^1. url:Why sum and product types? 2015-12-07
A function a->b?
The way I am reading what has been said is that it would give b^a, rather than a^b because a -> 1 should have 1^a == 1 possible states and 1->b should have b^1.