Closed roboguy13 closed 7 years ago
Is there a transformation rule that turns (\x -> (\y -> ...) b) a into (\x -> \y -> ...) a b? I'm introducing two lambdas with abstract calls and getting the former but I ultimately need the latter.
(\x -> (\y -> ...) b) a
(\x -> \y -> ...) a b
abstract
I figured it out: flip the order I make the abstract calls in and, before the second one, set the focus to the first lambda (with appFun).
appFun
Is there a transformation rule that turns
(\x -> (\y -> ...) b) a
into(\x -> \y -> ...) a b
? I'm introducing two lambdas withabstract
calls and getting the former but I ultimately need the latter.