Created
February 14, 2024 02:29
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Seemingly impossible functional program
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\import Data.Bool | |
\import Data.List | |
\import Logic.Meta | |
\import Paths | |
\func CoNat => | |
\Sigma (f : Nat -> Bool) | |
(\Pi (i : _) -> f i = false -> f (suc i) = false) | |
\where { | |
\func blt (n m : Nat) : Bool | |
| 0, suc m => true | |
| suc n, suc m => blt n m | |
| _, 0 => false | |
\func ofNat (n : Nat) : CoNat => (blt __ n, lemma __ n) \where { | |
\func lemma (m n : Nat) : blt m n = false -> blt (suc m) n = false | |
| 0, 0 => \lam p => p | |
| 0, suc n => contradiction | |
| suc m, 0 => \lam p => p | |
| suc m, suc n => lemma m n | |
} | |
\func inf : CoNat => (\lam _ => false, \lam _ x => x) | |
\func preview (n : CoNat) : List Bool => | |
\let f => n.1 \in f 0 :: f 1 :: f 2 :: f 3 :: f 4 :: nil | |
\func wtf => preview (ofNat 3) | |
-- Checks if p i for i ≤ n is true | |
-- if n is 0, then we only need to check p 0 | |
-- if n is suc n', recursively check for n' and make sure p n is true | |
\func find' (p : CoNat -> Bool) (n : Nat) : Bool \elim n | |
| 0 => p (ofNat 0) | |
| suc n' \as n => find' p n' and p (ofNat n) | |
\func find (p : CoNat -> Bool) : CoNat => | |
(find' p, \lam _ => pmap (__ and _)) | |
\func forall (p : CoNat -> Bool) : Bool => p (find p) | |
} |
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