defintional equality types

Question

If $a$ and $b$ are definitionally equal terms of type $A$ - i.e., $a$ and $b$ can be $\beta \eta$ reduced to identical terms - what follows about the structure of the identity type $a=_A b$? For example, are both $\mbox{refl}_a$ and $\mbox{refl}_b$ objects of type $a=_A b$? And are $\mbox{refl}_a$ and $\mbox{refl}_b$ themselves definitionally equal in virtue of $a$ and $b$ being definitionally equal?

Answer 1

Yes.

If $a \doteq b$ (by which I mean $a$ and $b$ are definitionally equal) then the types

• $a = b$
• $a = a$
• $b = b$

are all definitionally equal. Moreover, we have a definitional equality $\text{refl}_a \doteq \text{refl}_b$ in this type.

These follow from the "substitution rules" in section $1.3$ of Egbert Rijke's new book Introduction to Homotopy Type Theory. For completeness, here are the relevant rules:

$$ \frac{\Gamma \vdash a \doteq a' : A \quad \Gamma, x:A, \Delta \vdash B \text{ type}}{\Gamma, \Delta[a/x] \vdash B[a/x] \doteq B[a'/x] \text{ type}} $$

$$ \frac{\Gamma \vdash a \doteq a' : A \quad \Gamma, x:A, \Delta \vdash b:B}{\Gamma, \Delta[a/x] \vdash b[a/x] \doteq b[a'/x] : B[a/x]} $$

The first rule tells us that all of our types are definitionally equal. The second tells us that the terms $\text{refl}_a$ and $\text{refl}_b$ are definitionally equal.

If you're worried about the fact that a term seems to have multiple possible types, we show this is ok in exercise $1.1$ of this same book, which asks us to derive the "element conversion rule":

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

This is a kind of weird exercise (especially to make the first exercise in the book), and you can find a proof here. The key idea is to use the "varianble conversion rules"

$$ \frac{\Gamma \vdash A \doteq A' \text{ type} \quad \Gamma, x:A, \Delta \vdash \mathcal{J}}{\Gamma, x : A', \Delta \vdash \mathcal{J}} $$

which hold for any judgement $\mathcal{J}$.

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I hope this helps ^_^