4 ms·
`dup a b = {x0 x1}` will just substitute `a` by `x0` and `b` by `x1`, no matter where `a` and `b` are. So, if you apply that, verbatin, to: (Pair λx0(λy0(P
by LightMachine 5y ago
`dup a b = {x0 x1}` will just substitute `a` by `x0` and `b` by `x1`, no matter where `a` and `b` are. So, if you apply that, verbatin, to:
(Pair λx0(λy0(Pair a c)) λx1(λy1(Pair b d)))
You get:
(Pair λx0(λy0(Pair x0 c)) λx1(λy1(Pair x1 d)))
Does that make sense?