2

Exercise 1.2 in Introduction to Homotopy Type Theory by Egbert Rijke. https://arxiv.org/abs/2212.11082

Give a derivation for the following congruence rule for element conversion:

$$ \frac{\Gamma \vdash A \doteq A'\ \text{type}\quad \Gamma \vdash a \doteq b : A}{\Gamma \vdash a \doteq b : A'} $$

Really struggling with this, I can get

$$ \frac{\Gamma \vdash a \doteq b : A}{\Gamma \vdash a : A} \quad \text{and} \quad \frac{\Gamma \vdash a \doteq b : A}{\Gamma \vdash b : A} $$

from the formation rules. And applying the element conversion rule I can get

$$\Gamma \vdash a:A' \quad \text{and} \quad \Gamma \vdash b: A'$$

which seems like progress, but looking over all the rules formally I don't know where to go from here

Andrés Duarte
  • 21
  • 2
  • See https://math.stackexchange.com/questions/4058326/derivation-of-term-conversion-rule for the term conversion rule. The congruence rule follows the same idea, using the congruence of substitution. – Naïm Camille Favier Jun 28 '25 at 02:12
  • Thanks for pointing out the congruence of substitution. I can get to $$\frac{\Gamma \vdash a \doteq b : A \quad \Gamma, x:A \vdash x:A'}{\Gamma \vdash x[a/x] \doteq x[b/x] : A'[a/x]}$$ but not comfortable with how something like this reduces to $$\Gamma \vdash a \doteq b : A'$$ or maybe I should pick something else for what is $b$ in the book's description of the congruence. – Andrés Duarte Jun 28 '25 at 07:04
  • Just like in the term conversion case, $x[a/x] \equiv a$ and $x[b/x] \equiv b$, and furthermore $A'[a/x] \equiv A'$ since $x$ is fresh for $A'$. That last point is a bit easier to see using de Bruijn indices and explicit substitutions: the second premise is really $\Gamma, A \vdash 0 : A'[p]$, where $\Gamma, A \vdash p : \Gamma$ is the weakening substitution. – Naïm Camille Favier Jun 28 '25 at 11:31

0 Answers0