Chris Smith
Oct 6 · 2 min read
  1. Equivalently, the problem is that your function is not monotone in the definedness order. If f(⊥) = 1, then anything more defined than ⊥ — and that is everything at all — must map to something more defined than 1. (Since 1 is completely defined, that means they must map to 1 exactly. So if f(⊥) = 1, then f must be a constant function.)

    Chris Smith

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    Software engineer at Google, volunteer math and computer science teacher, author of the CodeWorld platform, amateur ring theorist, and Haskell enthusiast.

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