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There are several possible notions of proof relevance. Let us consider three similar situations: An element of a sum $\Sigma (x : A) . P(x)$ is a pair $(a, p)$ where $a : A$ and $p$ is a proof of $P(a)$. An element of $\Sigma (x : A) . \|P(x)\|$, where $\|{-}\|$ is propositional truncation, is a pair $(a, q)$ where $a : A$ and $q$ is an equivalence class ...


13

I recommend that everyone first read Andrej Bauer's answer, as he covers all the basics extremely well. I agree with everything he says in his answer. I humbly offer more comments, even though I know less on this topic than he does - but I was mentioned by name, as was my project. When I gave a talk about agda-categories, I explained one thing about it that ...


13

The difficulty is in making such a reduction compatible with all the other reductions involving transport/coe. From one perspective it is a “confluence” problem. It is unfortunate that in the community, so much was said about "decidability of degeneracies" that we began to give the impression that this was the hard part of the algorithm (since as ...


7

We are going to show that in MLTT with propositional truncation the type $$\textstyle \prod_{A:U_0}\prod_{B:U_0} (\|A\| \to A) \times (\|B\| \to B) \to (\|A + B\| \to A + B) $$ has no inhabitants. Assume it did. We shall work in a specific model of MLTT with propositional truncation, namely assemblies over number realizability. It is not too important what ...


4

No, there can't be such a surjection. Here's how to derive a contradiction, if there is a surjective map $f : A \to U_n$, where $A:U_m$. Since $m\leq n$, we can pull $f$ back along the embedding $U_m \to U_n$. This gives a surjective map $f' : A' \to U_m$, and it is not hard to see that the type $A'$ is equivalent to a type in $U_m$. Thus we have a ...


4

But it does follow. The types $$A = \prod_{(n,m:\textrm{Nat})}(\textrm{plus}\ n\ m) = (\textrm{plus}\ n\ m)$$ and $$B = \prod_{(n,m:\textrm{Nat})}(\textrm{plus}\ n\ m) = (\textrm{plus}\ m\ n)$$ are both contractible. Indeed, they are both inhabited and because $\mathrm{Nat}$ is a set, its identity type is a proposition, hence so are $A$ and $B$, as they are ...


3

Addressing the question in the title: $\mathsf{\lambda n\,m.\,refl}$ is not a proof of commutativity by definition because addition is not a constant function by definition. Of course, the commutativity proof can be shown to be propositionally equal (by a "dependent" equality, or path-over-path) to the $\mathsf{refl}$-returning one, by Andrej's ...


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