New answers tagged lambda-calculus
1
We can remove the dependence on what the values stored in the sequences are by assuming that we pass "the same thing" as the first two arguments. Regardless of what $b_1, b_2, b_3$ are:
$$([b_1b_2b_3]\,x)\,x = \lambda f. ((f\,x)\,x)\,x$$
This gives you an effective way to tell the difference between an empty sequence and a non-empty sequence :
$$(([]\left&...
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