I think this is an easy question: `fun <F, A, B...
# arrowb
Bob Glamm
05/23/2019, 2:05 PMI think this is an easy question:
fun <F, A, B> q(f: (a: A) -> B): (fa: F<A>) -> F<B> = ...Bob Glamm
05/23/2019, 2:05 PMfun <F, A, B> q(f: (a: A) -> B): (fa: F<A>) -> F<B> = ...