Created
September 4, 2019 19:42
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all' : (a -> Bool) -> List a -> Bool | |
all' f [] = True | |
all' f (x :: xs) = f x && all' f xs | |
allLemma : f x = True -> all' f xs = True -> all' f (x :: xs) = True | |
allLemma p1 p2 = rewrite p1 in rewrite p2 in Refl | |
allFiltered : (f : a -> Bool) -> (xs : List a) -> all' f (filter f xs) = True | |
allFiltered f [] = Refl | |
allFiltered f (x :: xs) = lemma (allFiltered f xs) (f x) Refl where | |
lemma : all' f (filter f xs) = True -> (b : Bool) -> b = f x -> all' f (filter f (x :: xs)) = True | |
lemma p False p2 = | |
rewrite sym p2 in p | |
lemma p True p2 = rewrite sym p2 in allLemma (sym p2) p |
There's a section at the bottom of http://docs.idris-lang.org/en/latest/tutorial/views.html explaining this, basically fx
is the name of the variable that will hold the proof that the expression being matched on equals the corresponding result for each matcher branch.
Oh my god, this is incredible! It's like Idris has the "inspect" pattern from Agda built right in! Thanks for showing me
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@clayrat Nice :)
I'm curious about the syntax in the with-rule though. What are
proof
andfx
?I originally thought to conduct the proof using the with-rule like this to just match on the result of
f x
, but this didn't appear to be refining the types. That is,f x
was still appearing in types and wasn't getting replaced withTrue
orFalse
in the respective branches, so theifThenElse
expressions weren't reducing.