CODEWITHSUNDEEP
coq
Kristian
Kristian

Reputation: 1737

Decidable equality statement with Set vs. Prop

When looking at results for types with decidable equality (especially in Eqdep_dec) there are some of the results that (for a type A) require

  forall x y : A, x = y \/ x <> y

whereas some require

  forall x y : A, {x = y} + {x <> y}

It is my impression that it is the last one that is referred to as decidable equality, but I am very much uncertain what the difference is. I know that x = y \/ x <> y in Prop and {x = y} + {x <> y} is in Set, and I can prove the first one from the second one but not the other way around. As far as I understand it, it is because I am not allowed to construct values of type Prop from values of type Set.

Can anyone tell what the difference between the two are? Are there some example of a type for which the first statement can be proved but not the second. Also, is it true that the version with {x = y} + {x <> y} is what is referred to as decidable equality?

Upvotes: 4

Views: 245

Answers (1)

Grant Jurgensen
Grant Jurgensen

Reputation: 179

You are correct that the latter definition, the one which lives in Set, is referred to as decidable equality.

Intuitively, we interpret objects in Set as programs, and objects in Prop as proofs. So the decidable equality type is the type of a function which takes any two elements of some type A and decides whether they are equal or unequal.

The other statement is slightly weaker. It describes the proposition that any two elements of A are either equal of unequal. Notably, we would not be able to inspect which outcome is the case for specific values of x and y, at least outside of case analysis within a proof. This is the result of the Prop elimination restriction that you alluded to (although you got it backwards: one is not allowed to construct values of sort Set/Type by eliminating/matching on an element of sort Prop).

Without adding axioms, the Prop universe is constructive, so I believe that there would not be any types A such that equality is undecidable but the propositional variant is provable. However, consider the scenario in which we make the Prop universe classical by adding the following axiom:

Axiom classic : forall P, P \/ ~P

This would make the propositional variant trivially provable for any type A, while the decidable equality may not be realizable.

Note that our axiom is a proposition. Intuitively, it makes sense that either a proposition or its negation must hold. If we hadn't made this a Prop (for example, if we axiomatized forall P, {P} + {~P}), then we would likely not accept the axiom, since it would instead be declaring the existence of a universal decision procedure.

That was a bit of a digression, but hopefully it demonstrated some differences in our interpretation of Props and Sets.

Upvotes: 6

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