Closed ghost closed 8 years ago
If we unconditionally provide an implicit Bimonad[Try] then we don't need separate Monad[Try] and Comonad[Try] values do we?
Bimonad[Try]
Monad[Try]
Comonad[Try]
@non Fixed/ready for review
:+1:
If we unconditionally provide an implicit
Bimonad[Try]
then we don't need separateMonad[Try]
andComonad[Try]
values do we?