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161 lines
4.8 KiB
Idris
161 lines
4.8 KiB
Idris
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module Data.Logic.Propositional
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%default total
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-- Idris uses intuitionistic logic, so it does not validate the
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-- principle of excluded middle (PEM) or similar theorems of classical
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-- logic. But it does validate certain relations among these
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-- propositions, and that's what's in this file.
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||| The principle of excluded middle
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PEM : Type -> Type
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PEM p = Either p (Not p)
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||| Double negation elimination
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DNE : Type -> Type
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DNE p = Not $ Not p -> p
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||| The consensus theorem (at least, the interesting part)
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Consensus : Type -> Type -> Type -> Type
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Consensus p q r = (q, r) -> Either (p, q) (Not p, r)
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||| Peirce's law
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Peirce : Type -> Type -> Type
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Peirce p q = ((p -> q) -> p) -> p
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||| Not sure if this one has a name, so call it Frege's law
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Frege : Type -> Type -> Type -> Type
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Frege p q r = (p -> r) -> (q -> r) -> ((p -> q) -> q) -> r
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||| The converse of contraposition: (p -> q) -> Not q -> Not p
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Transposition : Type -> Type -> Type
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Transposition p q = (Not q -> Not p) -> p -> q
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||| This isn't a good name.
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Switch : Type -> Type
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Switch p = (Not p -> p) -> p
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-- PEM and the others can't be proved outright, but it is possible to
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-- prove the double negations (DN) of all of them.
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||| The double negation of a proposition
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DN : Type -> Type
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DN p = Not $ Not p
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pemDN : DN $ PEM p
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pemDN f = f $ Right $ \x => f $ Left x
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dneDN : DN $ DNE p
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dneDN f = f $ \g => void $ g $ \x => f $ \_ => x
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consensusDN : DN $ Consensus p q r
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consensusDN f = f $ \(y, z) => Right (\x => f $ \(_, _) => Left (x, y), z)
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peirceDN : DN $ Peirce p q
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peirceDN f = f $ \g => g $ \x => void $ f $ \_ => x
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fregeDN : DN $ Frege p q r
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fregeDN f = f $ \g, h, i => h $ i $ \x => void $ f $ \_, _, _ => g x
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transpositionDN : DN $ Transposition p q
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transpositionDN f = f $ \g, x => void $ g (\y => f $ \_, _ => y) x
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switchDN : DN $ Switch p
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switchDN f = f $ \g => g $ \x => f $ \_ => x
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-- It's easy to prove all these theorems assuming PEM.
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dnePEM : PEM p -> DNE p
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dnePEM (Left l) _ = l
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dnePEM (Right r) f = void $ f r
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consensusPEM : PEM p -> Consensus p q r
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consensusPEM (Left l) (y, _) = Left (l, y)
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consensusPEM (Right r) (_, z) = Right (r, z)
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peircePEM : PEM p -> Peirce p q
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peircePEM (Left l) _ = l
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peircePEM (Right r) f = f $ \x => void $ r x
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fregePEM : PEM p -> Frege p q r
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fregePEM (Left l) f _ _ = f l
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fregePEM (Right r) _ g h = g $ h $ \x => void $ r x
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transpositionPEM : PEM p -> Transposition q p
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transpositionPEM (Left l) _ _ = l
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transpositionPEM (Right r) f x = void $ f r x
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switchPEM : PEM p -> Switch p
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switchPEM (Left l) _ = l
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switchPEM (Right r) f = f r
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-- It's trivial to prove these theorems assuming double negation
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-- elimination (DNE), since their double negations can all be proved.
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||| Eliminate double negations
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EDN : DN p -> DNE p -> p
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EDN f g = g f
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pemDNE : DNE $ PEM p -> PEM p
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pemDNE = EDN pemDN
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-- It's possible to prove the theorems assuming Peirce's law, but some
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-- thought must be given to choosing the right instances. Peirce's law
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-- is therefore weaker than PEM on an instance-by-instance basis, but
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-- all the instances together are equivalent.
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pemPeirce : Peirce (PEM p) Void -> PEM p
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pemPeirce f = f $ \g => Right $ \x => g $ Left x
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dnePeirce : Peirce p Void -> DNE p
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dnePeirce f g = f $ \h => void $ g h
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consensusPeirce : Peirce (Consensus p q r) Void -> Consensus p q r
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consensusPeirce f (y, z) =
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f (\g, (_, _) => Right (\x => g (\_ => Left (x, y)), z)) (y, z)
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fregePeirce : Peirce (Frege p q r) Void -> Frege p q r
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fregePeirce f g h i =
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f (\j, _, _, _ => h $ i $ \x => void $ j $ \_, _, _ => g x) g h i
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transpositionPeirce : Peirce (Transposition p q) Void -> Transposition p q
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transpositionPeirce f g x =
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f (\h, _, _ => void $ g (\y => h $ \_, _ => y) x) (\j, _ => g j x) x
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-- There are a variety of single axioms sufficient for deriving all of
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-- classical logic. The earliest of these is known as Nicod's axiom,
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-- but it is written using only nand operators and doesn't lend itself
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-- to Idris's type system. Meredith's axiom, on the other hand, is
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-- written using implication and negation.
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||| Meredith's axiom, sufficient for deriving all of classical logic
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Meredith : (p, q, r, s, t : Type) -> Type
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Meredith p q r s t =
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((((p -> q) -> Not r -> Not s) -> r) -> t) -> (t -> p) -> s -> p
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-- The Meredith axiom implies all of classical logic, and so in
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-- particular it implies PEM, and therefore cannot be proved in
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-- intuitionistic logic. As usual, however, its double negation can be
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-- proved.
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meredithDN : DN $ Meredith p q r s t
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meredithDN f =
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f $ \g, h, x =>
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h $ g $ \i =>
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void $ i
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(\y => void $ f $ \_, _, _ => y)
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(\z => f $ \_, _, _ => h $ g $ \_ => z)
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x
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-- Meredith can be proved assuming PEM, since the type system itself
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-- takes care of the rest.
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meredithPEM : PEM p -> Meredith p q r s t
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meredithPEM (Left l) _ _ _ = l
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meredithPEM (Right r) f g x =
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g $ f $ \h =>
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void $ h
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(\y => void $ r y)
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(\z => r $ g $ f (\_ => z))
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x
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