mirror of
https://github.com/idris-lang/Idris2.git
synced 2024-12-18 00:31:57 +03:00
21c6f4fb79
* [ breaking ] remove parsing of dangling binders It used to be the case that ``` ID : Type -> Type ID a = a test : ID (a : Type) -> a -> a test = \ a, x => x ``` and ``` head : List $ a -> Maybe a head [] = Nothing head (x :: _) = Just x ``` were accepted but these are now rejected because: * `ID (a : Type) -> a -> a` is parsed as `(ID (a : Type)) -> a -> a` * `List $ a -> Maybe a` is parsed as `List (a -> Maybe a)` Similarly if you want to use a lambda / rewrite / let expression as part of the last argument of an application, the use of `$` or parens is now mandatory. This should hopefully allow us to make progress on #1703
161 lines
4.8 KiB
Idris
161 lines
4.8 KiB
Idris
module Data.Logic.Propositional
|
|
|
|
%default total
|
|
|
|
-- Idris uses intuitionistic logic, so it does not validate the
|
|
-- principle of excluded middle (PEM) or similar theorems of classical
|
|
-- logic. But it does validate certain relations among these
|
|
-- propositions, and that's what's in this file.
|
|
|
|
||| The principle of excluded middle
|
|
PEM : Type -> Type
|
|
PEM p = Either p (Not p)
|
|
|
|
||| Double negation elimination
|
|
DNE : Type -> Type
|
|
DNE p = Not (Not p) -> p
|
|
|
|
||| The consensus theorem (at least, the interesting part)
|
|
Consensus : Type -> Type -> Type -> Type
|
|
Consensus p q r = (q, r) -> Either (p, q) (Not p, r)
|
|
|
|
||| Peirce's law
|
|
Peirce : Type -> Type -> Type
|
|
Peirce p q = ((p -> q) -> p) -> p
|
|
|
|
||| Not sure if this one has a name, so call it Frege's law
|
|
Frege : Type -> Type -> Type -> Type
|
|
Frege p q r = (p -> r) -> (q -> r) -> ((p -> q) -> q) -> r
|
|
|
|
||| The converse of contraposition: (p -> q) -> Not q -> Not p
|
|
Transposition : Type -> Type -> Type
|
|
Transposition p q = (Not q -> Not p) -> p -> q
|
|
|
|
||| This isn't a good name.
|
|
Switch : Type -> Type
|
|
Switch p = (Not p -> p) -> p
|
|
|
|
-- PEM and the others can't be proved outright, but it is possible to
|
|
-- prove the double negations (DN) of all of them.
|
|
|
|
||| The double negation of a proposition
|
|
DN : Type -> Type
|
|
DN p = Not $ Not p
|
|
|
|
pemDN : DN $ PEM p
|
|
pemDN f = f $ Right $ f . Left
|
|
|
|
dneDN : DN $ DNE p
|
|
dneDN f = f $ \g => void $ g $ \x => f $ \_ => x
|
|
|
|
consensusDN : DN $ Consensus p q r
|
|
consensusDN f = f $ \(y, z) => Right (\x => f $ \(_, _) => Left (x, y), z)
|
|
|
|
peirceDN : DN $ Peirce p q
|
|
peirceDN f = f $ \g => g $ \x => void $ f $ \_ => x
|
|
|
|
fregeDN : DN $ Frege p q r
|
|
fregeDN f = f $ \g, h, i => h $ i $ \x => void $ f $ \_, _, _ => g x
|
|
|
|
transpositionDN : DN $ Transposition p q
|
|
transpositionDN f = f $ \g, x => void $ g (\y => f $ \_, _ => y) x
|
|
|
|
switchDN : DN $ Switch p
|
|
switchDN f = f $ \g => g $ \x => f $ \_ => x
|
|
|
|
-- It's easy to prove all these theorems assuming PEM.
|
|
|
|
dnePEM : PEM p -> DNE p
|
|
dnePEM (Left l) _ = l
|
|
dnePEM (Right r) f = void $ f r
|
|
|
|
consensusPEM : PEM p -> Consensus p q r
|
|
consensusPEM (Left l) (y, _) = Left (l, y)
|
|
consensusPEM (Right r) (_, z) = Right (r, z)
|
|
|
|
peircePEM : PEM p -> Peirce p q
|
|
peircePEM (Left l) _ = l
|
|
peircePEM (Right r) f = f $ absurd . r
|
|
|
|
fregePEM : PEM p -> Frege p q r
|
|
fregePEM (Left l) f _ _ = f l
|
|
fregePEM (Right r) _ g h = g $ h $ absurd . r
|
|
|
|
transpositionPEM : PEM p -> Transposition q p
|
|
transpositionPEM (Left l) _ _ = l
|
|
transpositionPEM (Right r) f x = void $ f r x
|
|
|
|
switchPEM : PEM p -> Switch p
|
|
switchPEM (Left l) _ = l
|
|
switchPEM (Right r) f = f r
|
|
|
|
-- It's trivial to prove these theorems assuming double negation
|
|
-- elimination (DNE), since their double negations can all be proved.
|
|
|
|
||| Eliminate double negations
|
|
EDN : DN p -> DNE p -> p
|
|
EDN f g = g f
|
|
|
|
pemDNE : DNE (PEM p) -> PEM p
|
|
pemDNE = EDN pemDN
|
|
|
|
-- It's possible to prove the theorems assuming Peirce's law, but some
|
|
-- thought must be given to choosing the right instances. Peirce's law
|
|
-- is therefore weaker than PEM on an instance-by-instance basis, but
|
|
-- all the instances together are equivalent.
|
|
|
|
pemPeirce : Peirce (PEM p) Void -> PEM p
|
|
pemPeirce f = f $ \g => Right $ g . Left
|
|
|
|
dnePeirce : Peirce p Void -> DNE p
|
|
dnePeirce f g = f $ absurd . g
|
|
|
|
consensusPeirce : Peirce (Consensus p q r) Void -> Consensus p q r
|
|
consensusPeirce f (y, z) =
|
|
f (\g, (_, _) => Right (\x => g (\_ => Left (x, y)), z)) (y, z)
|
|
|
|
fregePeirce : Peirce (Frege p q r) Void -> Frege p q r
|
|
fregePeirce f g h i =
|
|
f (\j, _, _, _ => h $ i $ \x => void $ j $ \_, _, _ => g x) g h i
|
|
|
|
transpositionPeirce : Peirce (Transposition p q) Void -> Transposition p q
|
|
transpositionPeirce f g x =
|
|
f (\h, _, _ => void $ g (\y => h $ \_, _ => y) x) (\j, _ => g j x) x
|
|
|
|
-- There are a variety of single axioms sufficient for deriving all of
|
|
-- classical logic. The earliest of these is known as Nicod's axiom,
|
|
-- but it is written using only nand operators and doesn't lend itself
|
|
-- to Idris's type system. Meredith's axiom, on the other hand, is
|
|
-- written using implication and negation.
|
|
|
|
||| Meredith's axiom, sufficient for deriving all of classical logic
|
|
Meredith : (p, q, r, s, t : Type) -> Type
|
|
Meredith p q r s t =
|
|
((((p -> q) -> Not r -> Not s) -> r) -> t) -> (t -> p) -> s -> p
|
|
|
|
-- The Meredith axiom implies all of classical logic, and so in
|
|
-- particular it implies PEM, and therefore cannot be proved in
|
|
-- intuitionistic logic. As usual, however, its double negation can be
|
|
-- proved.
|
|
|
|
meredithDN : DN $ Meredith p q r s t
|
|
meredithDN f =
|
|
f $ \g, h, x =>
|
|
h $ g $ \i =>
|
|
void $ i
|
|
(\y => void $ f $ \_, _, _ => y)
|
|
(\z => f $ \_, _, _ => h $ g $ \_ => z)
|
|
x
|
|
|
|
-- Meredith can be proved assuming PEM, since the type system itself
|
|
-- takes care of the rest.
|
|
|
|
meredithPEM : PEM p -> Meredith p q r s t
|
|
meredithPEM (Left l) _ _ _ = l
|
|
meredithPEM (Right r) f g x =
|
|
g $ f $ \h =>
|
|
void $ h
|
|
(absurd . r)
|
|
(\z => r $ g $ f (\_ => z))
|
|
x
|