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The following Idris proof doesn’t typecheck.

hts : (s : Stream a) -> (head s :: tail s = s)
hts (x::xs) = Refl

The error I get is:

    Type mismatch between
            x :: Delay xs = x :: Delay xs
    and
            x :: Delay (tail (x :: Delay xs)) = x :: Delay xs

A similar proof for non-empty Vects typechecks just fine:

import Data.Vect
htv : (s : Vect (S k) a) -> (head s :: tail s = s)
htv (x::xs) = Refl

So I’m guessing the issue is in the laziness of Stream.

My working theory is that Idris doesn’t like simplifying whatever is inside of Delay, because it might get into an infinite loop that way. However, I want to coerce Idris to dip its toes in anyway, as the definition of Prelude.Stream.tail guarantees that the LHS would then reduce to x :: Delay xs, completing my proof.

Is my suspicion correct? And can I fix the proof somehow?

Lynn
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1 Answers1

2

Yes, it can be done. I used an auxiliary congruence lemma: 

%default total

consCong : {xs, ys : Stream a} -> (x : a) -> xs = ys -> x :: xs = x :: ys
consCong _ Refl = Refl

to prove the main lemma:

hts : (s : Stream a) -> (head s :: tail s = s)
hts (x :: xs) = consCong _ $ Refl
Anton Trunov
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