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by LightMachine
1593 days ago
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`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? |
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