Skip to main content
Proof Assistants Beta

You are not logged in. Your edit will be placed in a queue until it is peer reviewed.

We welcome edits that make the post easier to understand and more valuable for readers. Because community members review edits, please try to make the post substantially better than how you found it, for example, by fixing grammar or adding additional resources and hyperlinks.

4
  • $\begingroup$ If you have $x + y = y + x$ judgmentally with $x$ and $y$ variables, this easily contradicts termination of MLTT computation. Without that you'll need a more subtle argument. Note that the free monoid over a finite decidable type can be encoded using a clever NbE trick. $\endgroup$
    – cody
    Commented Aug 9, 2023 at 18:23
  • $\begingroup$ @cody Can you elaborate on the termination argument? Why can't they both reduce to a common expression? $\endgroup$
    – Trebor ♦
    Commented Aug 10, 2023 at 3:45
  • $\begingroup$ Because any normal form of the LHS would have to be $\alpha$ equivalent to itself, with the variables $x$ and $y$ swapped (since reduction in MLTT commutes with substitution). It's not too hard to see that such a term cannot exist. $\endgroup$
    – cody
    Commented Aug 10, 2023 at 15:49
  • $\begingroup$ @cody Ah, nice argument. I think that's already worth an answer (perhaps with some generalizations). $\endgroup$
    – Trebor ♦
    Commented Aug 10, 2023 at 15:58