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I attempted to compare a String and String, expecting True.

Idris> String == String
Can't find implementation for Eq Type

Then I expected False when comparing a String to a Bool.

Idris> String /= Bool
Can't find implementation for Eq Type

Am I missing an import?

Kevin Meredith
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2 Answers2

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You can't as it would break parametricity, which we have in Idris. We can't pattern match on types. But this would be necessary to write the Eq implementation, for example:

{- Doesn't work!
eqNat : Type -> Bool
eqNat Nat = True
eqNat _ = False -}

Also, if one could pattern match on types, they would be needed in the run-time. Right now types get erased when compiling.

xash
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  • Is parametric desired for being able to have the compiler/interface derive the free terms? – Nikolaj-K Apr 11 '16 at 19:04
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    If in `A == B` `A` and `B` are in canonical form and belong to some closed universe, then you don't need to break parametricity to be able to check them for equality (and you can enclose the whole `Type` using induction-recursion and some form of generic programming like they did in Observational Type Theory). [Here](https://github.com/effectfully/random-stuff/blob/master/EqType.agda) is some code. Though, operating on terms in canonical form is mostly useless, especially when you have a non-parametric universe that offers explicit codes for types. – effectfully Apr 24 '16 at 08:23
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Just to add some simple examples to the above: types can't be pattern matched on, but there's a two parameter type constructor for propositional equality, described in the documentation section on Theorem Proving. Notice that the only constructor, Refl, makes only values of type (=) x x, where both type parameters are the same. (this is ≡ in Agda)

So this will typecheck:

twoPlusTwoEqFour : 2 + 2 = 4
twoPlusTwoEqFour = Refl

so will this:

stringEqString : String = String
stringEqString = Refl

but not this:

stringEqInt : String = Int
stringEqInt = Refl

-- type error: Type mismatch between String and Int

and this needs extra work to prove, because addition is defined by recursion on the left argument, and n + 0 can't be reduced further:

proof : n = n + 0
brainrape
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