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{{draft task|Logic}} {{omit from|C}} {{omit from|C++}} {{Omit From|D}} {{Omit From|E}} {{Omit From|Ruby}} {{Omit From|ALGOL 68}} {{omit from|Java}} {{omit from|MUMPS|Not a typed language}} {{omit from|Python}} {{omit from|M4}} {{omit from|Metafont}} {{omit from|Smalltalk}} {{omit from|Fortran}} {{omit from|AWK}} {{omit from|Objective-C}}
This task only makes sense for [[wp:dependently-typed language|dependently-typed languages]] and [[wp:proof assistant|proof assistants]], or for languages with a type system strong enough to emulate certain dependent types. It does ''not'' ask you to implement a theorem prover yourself.
In the following task the word "define" implies the need to build the system of [[wp:Peano_axioms|Peano axioms]] using the language itself, that is a way to construct natural and even natural numbers in the [[wp:Canonical_form|canonical forms]], as well as a definition of the rules of addition and a way to construct all other acceptable terms. The word "prove" means that some form of [[wp:Unification_(computer_science)|logical unification]] is used (i.e., it requires a type checker in the case of languages with dependent types and a verifying algorithm in the case of proof assistants). Thus, the metatheory of a language must be expressive enough to allow embedding of the Peano axioms and the opportunity to carry out [[wp:Constructive_proof|constructive proofs]]. Examples of the ''trusted'' mathematical metatheories can be given: [[wp:System_F|SystemF]] for Haskell, [[wp:Intuitionistic_type_theory|MLTT]] for Agda, [[wp:Calculus_of_constructions|CoC]]/[[wp:Calculus_of_inductive_constructions|CoIC]] for Coq.
'''Task''':
To illustrate the possibility of type formation and type introduction:
Define a countably infinite set of natural numbers {0, 1, 2, 3, ...}.
Define a countably infinite set of even natural numbers {0, 2, 4, 6, ...} within the previously defined set of natural numbers.
Define a countably infinite set of odd natural numbers {1, 3, 5, 7, ...} within the previously defined set of natural numbers.
To illustrate the possibility of type elimination:
Define the addition on natural numbers.
To demonstrate constructive proofs:
Prove that the addition of ''any'' two even numbers is even.
Prove that the addition is ''always'' associative.
Prove that the addition is ''always'' commutative.
Try to prove that the addition of ''any'' two even numbers is odd (it should be rejected).
To demonstrate the ability of disproofs:
Prove that the addition of ''any'' two even numbers cannot be odd.
Try to prove that the addition of ''any'' two even numbers cannot be even (it should be rejected).
- and 4. can't be done using a simple number enumeration since there is a countable many natural numbers which is quantified in propositions.
ACL2
3.1. using built-in natural numbers:
(thm (implies (and (evenp x) (evenp y))
(evenp (+ x y))))
Agda
module PeanoArithmetic where
-- 1.1. The natural numbers.
--
-- ℕ-formation: ℕ is set.
--
-- ℕ-introduction: 0 ∈ ℕ,
-- a ∈ ℕ | (1 + a) ∈ ℕ.
--
data ℕ : Set where
zero : ℕ
1+_ : ℕ → ℕ
{-# BUILTIN NATURAL ℕ #-}
{-# BUILTIN ZERO zero #-}
{-# BUILTIN SUC 1+_ #-}
-- 2.1. The rule of addition.
--
-- via ℕ-elimination.
--
infixl 6 _+_
_+_ : ℕ → ℕ → ℕ
0 + n = n
1+ m + n = 1+ (m + n)
-- 1.2. The even natural numbers.
--
data 2×ℕ : ℕ → Set where
zero : 2×ℕ 0
2+_ : {m : ℕ} → 2×ℕ m → 2×ℕ (2 + m)
-- 1.3. The odd natural numbers.
--
data 2×ℕ+1 : ℕ → Set where
one : 2×ℕ+1 1
2+_ : {m : ℕ} → 2×ℕ+1 m → 2×ℕ+1 (2 + m)
-- 3.1. Sum of any two even numbers is even.
--
-- This function takes any two even numbers and returns their sum as an even
-- number, this is the type, i.e. logical proposition, algorithm itself is a
-- proof which builds a required term of a given (inhabited) type, and the
-- typechecker performs that proof (by unification, so that this is a form of
-- compile-time verification).
--
even+even≡even : {m n : ℕ} → 2×ℕ m → 2×ℕ n → 2×ℕ (m + n)
even+even≡even zero n = n
even+even≡even (2+ m) n = 2+ (even+even≡even m n)
-- The identity type for ℕ (for propositional equality).
--
infix 4 _≡_
data _≡_ (m : ℕ) : ℕ → Set where
refl : m ≡ m
sym : {m n : ℕ} → m ≡ n → n ≡ m
sym refl = refl
trans : {m n p : ℕ} → m ≡ n → n ≡ p → m ≡ p
trans refl n≡p = n≡p
-- refl, sym and trans forms an equivalence relation.
cong : {m n : ℕ} → m ≡ n → 1 + m ≡ 1 + n
cong refl = refl
-- 3.2.1. Direct proof of the associativity of addition using propositional
-- equality.
--
+-associative : (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+-associative 0 _ _ = refl
+-associative (1+ m) n p = cong (+-associative m n p)
-- Proof _of_ mathematical induction on the natural numbers.
--
-- P 0, ∀ x. P x → P (1 + x) | ∀ x. P x.
--
ind : (P : ℕ → Set) → P 0 → ((m : ℕ) → P m → P (1 + m)) → (m : ℕ) → P m
ind _ P₀ _ 0 = P₀
ind P P₀ next (1+ n) = next n (ind P P₀ next n)
-- 3.2.2. The associativity of addition by induction (with propositional
-- equality, again).
--
+-associative′ : (m n p : ℕ) → (m + n) + p ≡ m + (n + p)
+-associative′ m n p = ind P P₀ is m
where
P : ℕ → Set
P m = m + n + p ≡ m + (n + p)
P₀ : P 0
P₀ = refl
is : (m : ℕ) → P m → P (1 + m)
is _ Pi = cong Pi
-- Syntactic sugar for equational reasoning (we don't use preorders here).
infix 4 _≋_
data _≋_ (m n : ℕ) : Set where
refl : m ≡ n → m ≋ n
infix 1 begin_
begin_ : {m n : ℕ} → m ≋ n → m ≡ n
begin refl m≡n = m≡n
infixr 2 _~⟨_⟩_
_~⟨_⟩_ : (m : ℕ){n p : ℕ} → m ≡ n → n ≋ p → m ≋ p
_ ~⟨ m≡n ⟩ refl n≡p = refl (trans m≡n n≡p)
infix 2 _∎
_∎ : (m : ℕ) → m ≋ m
_∎ _ = refl refl
-- Some helper proofs.
m+0≡m : (m : ℕ) → m + 0 ≡ m
m+0≡m 0 = refl
m+0≡m (1+ m) = cong (m+0≡m m)
m+1+n≡1+m+n : (m n : ℕ) → m + (1 + n) ≡ 1 + (m + n)
m+1+n≡1+m+n 0 n = refl
m+1+n≡1+m+n (1+ m) n = cong (m+1+n≡1+m+n m n)
-- 3.3. The commutativity of addition using equational reasoning.
--
+-commutative : (m n : ℕ) → m + n ≡ n + m
+-commutative 0 n = sym (m+0≡m n)
+-commutative (1+ m) n =
begin
1+ m + n ~⟨ refl ⟩
1+ (m + n) ~⟨ cong (+-commutative m n) ⟩
1+ (n + m) ~⟨ sym (m+1+n≡1+m+n n m) ⟩
n + 1+ m
∎
-- 3.4.
--
even+even≡odd : {m n : ℕ} → 2×ℕ m → 2×ℕ n → 2×ℕ+1 (m + n)
even+even≡odd zero zero = {!!}
even+even≡odd _ _ = {!!}
-- ^
-- That gives
--
-- ?0 : 2×ℕ+1 (zero + zero)
-- ?1 : 2×ℕ+1 (.m + .n)
--
-- but 2×ℕ+1 (zero + zero) = 2×ℕ+1 0 which is uninhabited, so that this proof
-- can not be writen.
--
-- The absurd (obviously uninhabited) type.
--
-- ⊥-introduction is empty.
--
data ⊥ : Set where
-- The negation of a proposition.
--
infix 6 ¬_
¬_ : Set → Set
¬ A = A → ⊥
-- 4.1. Disproof or proof by contradiction.
--
-- To disprove even+even≡odd we assume that even+even≡odd and derive
-- absurdity, i.e. uninhabited type.
--
even+even≢odd : {m n : ℕ} → 2×ℕ m → 2×ℕ n → ¬ 2×ℕ+1 (m + n)
even+even≢odd zero zero ()
even+even≢odd zero (2+ n) (2+ m+n) = even+even≢odd zero n m+n
even+even≢odd (2+ m) n (2+ m+n) = even+even≢odd m n m+n
-- 4.2.
--
-- even+even≢even : {m n : ℕ} → 2×ℕ m → 2×ℕ n → ¬ 2×ℕ (m + n)
-- even+even≢even zero zero ()
-- ^
-- rejected with the following message:
--
-- 2×ℕ zero should be empty, but the following constructor patterns
-- are valid:
-- zero
-- when checking that the clause even+even≢even zero zero () has type
-- {m n : ℕ} → 2×ℕ m → 2×ℕ n → ¬ 2×ℕ (m + n)
--
Coq
1.1., 1.2., 2.1. and 3.1.:
Inductive nat : Set :=
| O : nat
| S : nat -> nat.
Fixpoint plus (n m:nat) {struct n} : nat :=
match n with
| O => m
| S p => S (p + m)
end
where "n + m" := (plus n m) : nat_scope.
Inductive even : nat -> Set :=
| even_O : even O
| even_SSn : forall n:nat,
even n -> even (S (S n)).
Theorem even_plus_even : forall n m:nat,
even n -> even m -> even (n + m).
Proof.
intros n m H H0.
elim H.
trivial.
intros.
simpl.
case even_SSn.
intros.
apply even_SSn; assumption.
assumption.
Qed.
Haskell
Using [http://www.haskell.org/haskellwiki/GADT GADTs] and [http://www.haskell.org/haskellwiki/GHC/Type_families type families] it is possible to write a partial adaptation of the Agda version:
{-# LANGUAGE TypeOperators, TypeFamilies, GADTs #-}
module PeanoArithmetic where
-- 1.1. Natural numbers.
data Z = Z
data S m = S m
-- 2.1. Addition.
infixl 6 :+
type family x :+ y
type instance Z :+ n = n
type instance S m :+ n = S (m :+ n)
-- 1.2. Even natural numbers.
data En :: * -> * where
Ez :: En Z
Es :: En m -> En (S (S m))
-- 1.3. Odd natural numbers.
data On :: * -> * where
Oo :: On (S Z)
Os :: On m -> On (S (S m))
-- 3.1. Sum of any two even numbers is even.
sum_of_even_is_even :: En m -> En n -> En (m :+ n)
sum_of_even_is_even Ez n = n
sum_of_even_is_even (Es m) n = Es $ sum_of_even_is_even m n
-- The identity type for natural numbers.
infix 4 :=
data (:=) m :: * -> * where
Refl :: m := m
sym :: m := n -> n := m
sym Refl = Refl
trans :: m := n -> n := p -> m := p
trans Refl np = np
cong :: m := n -> S m := S n
cong Refl = Refl
-- 3.2. Associativity of addition (via propositional equality).
class AssocAdd m where
proof :: m -> n -> p -> (m :+ n) :+ p := m :+ (n :+ p)
instance AssocAdd Z where
proof Z _ _ = Refl
instance AssocAdd m => AssocAdd (S m) where
proof (S m) n p = cong $ proof m n p
-- Induction, associativity of addition by induction, equational reasoning and
-- commutativity of addition is too tricky.
-- 3.4. Bad proof.
sum_of_even_is_odd :: En m -> En n -> On (m :+ n)
-- ^
-- Сan not be written totally:
--
sum_of_even_is_odd Ez Ez = undefined
-- ^
-- then, in GHCi:
--
-- *PeanoArithmetic> :t sum_of_even_is_odd Ez Ez
-- sum_of_even_is_odd Ez Ez :: On (Z :+ Z)
-- *PeanoArithmetic> :t undefined :: On (Z :+ Z)
-- undefined :: On (Z :+ Z) :: On Z
-- *PeanoArithmetic> :t sum_of_even_is_odd Ez Ez :: On Z
-- sum_of_even_is_odd Ez Ez :: On Z :: On Z
-- *PeanoArithmetic> :t Oo
-- Oo :: On (S Z)
-- *PeanoArithmetic> :t Os Oo
-- Os Oo :: On (S (S (S Z)))
-- *PeanoArithmetic> :t Os (Os Oo)
-- Os (Os Oo) :: On (S (S (S (S (S Z)))))
--
-- so that sum_of_even_is_odd Ez Ez :: On Z, but On Z is empty, it is impossible
-- to write such a proof.
--
-- Uninhabited type.
data Bot
-- Negation.
type Not a = a -> Bot
-- 4.1. Disproof.
sum_of_even_is_not_odd :: En m -> En n -> Not (On (m :+ n))
sum_of_even_is_not_odd Ez (Es n) (Os mn) = sum_of_even_is_not_odd Ez n mn
sum_of_even_is_not_odd (Es m) n (Os mn) = sum_of_even_is_not_odd m n mn
sum_of_even_is_not_odd Ez Ez _ =
error "impossible happened in sum_of_even_is_not_odd!"
-- ^
-- partial, however, we know that Ez :: En Z, Z + Z = Z and On Z is
-- uninhabited, so that this clause is unreachable.
--
-- Also, GHC complains:
--
-- Warning: Pattern match(es) are non-exhaustive
-- In an equation for `sum_of_even_is_not_odd':
-- Patterns not matched:
-- Ez (Es _) Oo
-- (Es _) _ Oo
--
-- and can't find that this clauses is unreachable too, since this isn't type check:
--
-- sum_of_even_is_not_odd Ez (Es _) Oo = undefined
-- sum_of_even_is_not_odd (Es _) _ Oo = undefined
--
-- 4.2. Bad disproof.
sum_of_even_is_not_even :: En m -> En n -> Not (En (m :+ n))
--
-- Starting from a partial definition:
--
sum_of_even_is_not_even Ez Ez _ = undefined
--
-- we can show that it can not be rewritten totally:
--
-- *PeanoArithmetic> :t sum_of_even_is_not_even Ez Ez
-- sum_of_even_is_not_even Ez Ez :: Not (En (Z :+ Z))
-- *PeanoArithmetic> :t sum_of_even_is_not_even Ez Ez :: Not (En Z)
-- sum_of_even_is_not_even Ez Ez :: Not (En Z) :: Not (En Z)
-- *PeanoArithmetic> :t sum_of_even_is_not_even Ez Ez :: En Z -> Bot
-- sum_of_even_is_not_even Ez Ez :: En Z -> Bot :: En Z -> Bot
-- *PeanoArithmetic> :t Ez
-- Ez :: En Z
-- *PeanoArithmetic> :t (sum_of_even_is_not_even Ez Ez :: En Z -> Bot) Ez
-- (sum_of_even_is_not_even Ez Ez :: En Z -> Bot) Ez :: Bot
--
-- since we have a "citizen" of an uninhabited type here (contradiction!).
--
See also [[Proof/Haskell]] for implementation of a small theorem prover.
Idris
Idris supports two types of proofs: using tactics, like Coq, or just write functions like Agda. These ways can be combined.
module Proof
-- Enable terminator checker by default
%default total
-- 1.1 Natural numbers
%elim
data MyNat
= Z
| S MyNat
-- 1.2 Even naturals
%elim
data EvNat : MyNat -> Type where
EvO : EvNat Z
EvSS : EvNat x -> EvNat (S (S x))
-- 1.3 Odd naturals
%elim
data OddNat : MyNat -> Type where
Odd1 : OddNat (S Z)
OddSS : OddNat x -> OddNat (S (S x))
-- 2.1 addition
infixl 4 :+
(:+) : MyNat -> MyNat -> MyNat
(:+) Z b = b
(:+) (S a) b = S (a :+ b)
-- 3.1, Prove that the addition of any two even numbers is even.
evensPlus1 : {a : MyNat} -> {b : MyNat} -> (EvNat a) -> (EvNat b) -> (EvNat (a :+ b))
evensPlus1 ea eb = ?proof31
congS : {a : MyNat} -> {b : MyNat} -> (a = b) -> (S a = S b)
congS refl = refl
evensPlus2 : {a : MyNat} -> {b : MyNat} -> (EvNat a) ->(EvNat b) -> (EvNat (a :+ b))
evensPlus2 EvO eb = eb
evensPlus2 {a=(S (S a))} (EvSS ea) eb = EvSS (evensPlus2 ea eb)
-- 3.2 Prove that the addition is always associative.
plusAssoc : (a : MyNat) -> (b : MyNat) -> (c : MyNat) -> (a :+ b) :+ c = a :+ (b :+ c)
plusAssoc Z b c = refl
plusAssoc (S a) b c = congS (plusAssoc a b c)
-- 3.3 Prove that the addition is always commutative.
plus0_r : (a : MyNat) -> a :+ Z = a
plus0_r Z = refl
plus0_r (S a) = congS (plus0_r a)
plusS_r : (a : MyNat) -> (b : MyNat) -> a :+ S b = S (a :+ b)
plusS_r Z b = refl
plusS_r (S a) b = congS (plusS_r a b)
plusComm : (a : MyNat) -> (b : MyNat) -> a :+ b = b :+ a
plusComm a b = ?proof33
-- 4.1 Prove that the addition of any two even numbers cannot be odd.
evenNotOdd : (ea : EvNat a) -> (oa : OddNat a) -> _|_
evenNotOdd (EvSS e) (OddSS o) = evenNotOdd e o
evensPlusNotOdd : (ea : EvNat a) -> (eb : EvNat b) -> (OddNat (a :+ b)) -> _|_
evensPlusNotOdd EvO (EvSS eb) (OddSS ob) = evenNotOdd eb ob
evensPlusNotOdd (EvSS y) EvO oab = ?epno_so
evensPlusNotOdd (EvSS y) (EvSS z) oab = ?epno_ss
---------- Proofs ----------
Proof.proof31 = proof
intro a
intro b
intro ea
intro eb
induction ea
compute
exact eb
intro x
intro ex
intro exh
exact (EvSS exh)
Proof.proof33 = proof
intros
induction a
compute
rewrite sym (plus0_r b)
trivial
intro a'
intro ha'
compute
rewrite sym (plusS_r b a')
exact (congS ha')
Proof.epno_ss = proof
intro x
intro ex
intro y
intro ey
compute
rewrite sym ( plusS_r x (S y))
rewrite sym ( plusS_r x y)
intro os4xy
let es4xy = EvSS(EvSS (evensPlus1 ex ey))
exact evenNotOdd es4xy os4xy
Proof.epno_so = proof
intro x
intro ex
rewrite sym (plus0_r x)
compute
rewrite sym (plus0_r x)
intro ossx
exact evenNotOdd (EvSS ex) ossx
J
A Peano number needs a zero, a mechanism for distinguishing equality from its absence, a way of getting a successor to a number, and a system of induction.
We know that a computer can never fully implement peano numbers because a computer can only represent a finite number of distinct values (if necessary, Ackerman's function can be used to illustrate the existence of this limitation). So our implementation of Peano Numbers will represent these mechanisms rather than be a complete implementation of these mechanisms.
So, these can be our definitions:
context=:3 :0
if. 0 = L. y do. context (,: ; ]) y return. end.
kernel=. > {: y
symbols=. (#$kernel) {. > {. y
order=. /: symbols
(symbols /: order); order |: kernel
)
symbols=: >@{.
kernel=: >@{:
zero=: context0x
monadic=: (1 :'[:context u@symbols; u&kernel')( :[:)
dyadic=: (1 :'[:context ,&symbols; u&kernel')([: :)
successor=: +&1x monadic
equals=: -:&kernel
all=: i. >. %: kernel successor@$: ::] zero
is_member_of=: e.&(-. -.&all)&,&kernel ([: :)
exists_in=: 1&e.@e.&kernel ([: :)
not=: -. :[:
induction=:4 :0
3 :'(y)=:context ?~#all'&.>;:x
assert (#all) > #;._1 LF,y
assert 0 = # y -.&;: LF,defined,x
assert 0!:3 y
)
addition=: +/ dyadic
isN=: ([ assert@is_member_of&(context all))
even=: [: context kernel@addition~"0@,@kernel :[: @isN
odd=: successor@even
defined=: '(zero not exists_in odd successor equals is_member_of addition even)'
Here, '''even''' is a function which, given a natural number, produces a corresponding even natural number. '''odd''' is a similar function which gives us odd numbers.
And, '''induction''' is a verb where the left argument lists values which represent any natural number and the right argument represents an expression to be considered. If induction succeeds, the expression is true for all natural numbers.
Meanwhile '''-.''' is J's implementation of "logical negation" -- a function which has been grafted onto boolean algebra as a convenience for people working with logical systems. ("Grafted on" here means: [a] logical negation was not originally a part of the definition of boolean algebra, and [b] logical negation cannot be a valid part of an infinite set of systems that were valid boolean algebras before someone decided that logical negation should be a part of boolean algebra.)
Also '''is_member_of''' represents universal set membership, '''exists_in''' represents the existential quantifier.
Note that we do not have to prove that our definitions are correct -- they are our axioms that we use in our proof.
Thus, this is proof that
all natural numbers are even, and
addition is associative, and
addition is commutative, and
the sum of two even numbers can never be odd.
'A B C' induction 0 :0
((even A) addition (even B)) is_member_of (even C)
((A addition B) addition C) equals (A addition (B addition C))
(A addition B) equals (B addition A)
not ((even A) addition (even B)) exists_in (odd C)
)
Meanwhile, here is how the invalid proofs fail:
'A B C' induction '((even A) addition (even B)) is_member_of (odd C)'
|assertion failure: assert
and
'A B C' induction 'not ((even A) addition (even B)) is_member_of (even C)'
|assertion failure: assert
As an aside, note that peano numbers show us that numbers can represent recursive processes.
OCaml
Using GADT, we can port the Coq version to OCaml.
type zero = Zero
type 'a succ = Succ
(* 1.1 *)
type _ nat =
| Zero : zero nat
| Succ : 'a nat -> 'a succ nat
(* 1.2 *)
type _ even =
| Even_zero : zero even
| Even_ss : 'a even -> 'a succ succ even
(* 2.1: define the addition relation *)
type (_, _, _) plus =
| Plus_zero : ('a, zero, 'a) plus
| Plus_succ : ('a, 'b, 'c) plus -> ('a, 'b succ, 'c succ) plus
(* 3.1 *)
(* Define the property: there exists a number 'sum that is the sum of 'a and 'b and is even. *)
type ('a, 'b) exists_plus_even = Exists_plus_even : ('a, 'b, 'sum) plus * 'sum even -> ('a, 'b) exists_plus_even
(* The proof that if a and b are even, there exists sum that is the sum of a and b that is even.
This is a valid proof since it terminated, and it is total (no assert false, and the exhaustiveness
test of pattern matching doesn't generate a warning) *)
let rec even_plus_even : type a b. a even -> b even -> (a, b) exists_plus_even =
fun a_even b_even ->
match b_even with
| Even_zero ->
Exists_plus_even (Plus_zero, a_even)
| Even_ss b_even' ->
let Exists_plus_even (plus, even) = even_plus_even a_even b_even' in
Exists_plus_even (Plus_succ (Plus_succ plus), Even_ss even)
(* 3.3 *)
(* 0 + n = n *)
let rec plus_zero : type a. a nat -> (zero, a, a) plus = function
| Zero -> Plus_zero
| Succ a -> Plus_succ (plus_zero a)
(* a + b = n => (a + 1) + b = (n + 1) *)
let rec plus_succ_left : type a b sum. (a, b, sum) plus -> (a succ, b, sum succ) plus =
function
| Plus_zero -> Plus_zero
| Plus_succ plus -> Plus_succ (plus_succ_left plus)
(* a + b = n => b + a = n *)
let rec plus_commutative : type a b sum. a nat -> (a, b, sum) plus -> (b, a, sum) plus =
fun a plus ->
match plus with
| Plus_zero -> plus_zero a
| Plus_succ plus' ->
plus_succ_left (plus_commutative a plus')
Omega
data Even :: Nat ~> *0 where
EZ:: Even Z
ES:: Even n -> Even (S (S n))
plus:: Nat ~> Nat ~> Nat
{plus Z m} = m
{plus (S n) m} = S {plus n m}
even_plus:: Even m -> Even n -> Even {plus m n}
even_plus EZ en = en
even_plus (ES em) en = ES (even_plus em en)
Phix
Silly text-based version, just for fun.
Obviously 1.1/1.2/1.3 are (loosely) eumlated by the strings "int"/"even"/"odd", and addition by axioms[4].
Clearly 3.2 and 3.3 are not attempted, and I'm not sure which of 3.4/4.1/4.2 the last test is closest to, or for that matter what the difference between them is supposed to be.
constant axioms = {{"even+1","odd"},
{"even","2*int"},
{"2*int+2*int","2*(int+int)"},
{"(int+int)","int"},
{"2*int","even"},
{"odd","2*int+1"}}
procedure proof(string a, b)
-- try to convert a into b using only the above axioms
string w = a
sequence seen = {} -- (avoid infinite loops)
while not find(w,seen) do
seen = append(seen,w)
?w
if w=b then exit end if
integer hit = 0
for i=1 to length(axioms) do
string {r,s} = axioms[i]
integer k = match(r,w)
if k then
w[k..k+length(r)-1] = s
hit = i
exit
end if
end for
if hit=0 then exit end if
puts(1,"== ")
end while
?{a,b,iff(w=b?"true","false")}
end procedure
proof("even+even","even")
proof("even+1","odd")
--bad proofs:
proof("int","even")
proof("even+even","odd")
{{out}}
"even+even"
== "2*int+even"
== "2*int+2*int"
== "2*(int+int)"
== "2*int"
== "even"
{"even+even","even","true"}
"even+1"
== "odd"
{"even+1","odd","true"}
"int"
{"int","even","false"}
"even+even"
== "2*int+even"
== "2*int+2*int"
== "2*(int+int)"
== "2*int"
== "even"
== {"even+even","odd","false"}
Racket
Via #lang cur. Currently (August 29th, 2019), I am using a development version, which could be installed by the following commands:
raco pkg install https://github.com/stchang/macrotypes.git?path=macrotypes-lib#cur
raco pkg install https://github.com/stchang/macrotypes.git?path=turnstile-lib#cur
raco pkg install https://github.com/wilbowma/cur.git?path=cur-lib#turnstile-core
#lang cur
(require rackunit
cur/stdlib/equality
cur/stdlib/sugar
cur/ntac/base
cur/ntac/standard
cur/ntac/rewrite)
;; Task 1.1
(data nat : 0 Type
[O : nat]
[S : (-> nat nat)])
(define-syntax #%datum
(syntax-parser
[(_ . n:exact-nonnegative-integer)
#:when (zero? (syntax-e #'n))
#'O]
[(_ . n:exact-nonnegative-integer)
#`(S (#%datum . #,(- (syntax-e #'n) 1)))]))
(check-equal? (S (S (S (S O)))) 4)
;; Task 1.2
(data even : 0 (-> nat Type)
[even-O : (even 0)]
[even-SS : (forall [n : nat] (-> (even n) (even (S (S n)))))])
;; Task 1.3
(data odd : 0 (-> nat Type)
[odd-O : (odd 1)]
[odd-SS : (forall [n : nat] (-> (odd n) (odd (S (S n)))))])
;; Task 2.1
(define/rec/match + : nat [m : nat] -> nat
[O => m]
[(S n-1) => (S (+ n-1 m))])
(check-equal? (+ 2 3) 5)
;; Task 3.1
(define-theorem even-plus-even-is-even
(∀ [n : nat] [m : nat] (-> (even n) (even m) (even (+ n m))))
(by-intros n m Hn Hm)
(by-induction Hn #:as [() (n* IHn)])
;; subgoal 1
simpl
by-assumption
;; subgoal 2
(by-apply even-SS)
by-assumption)
;; Task 3.2
(define-theorem addition-assoc
(∀ [a : nat] [b : nat] [c : nat] (== nat (+ (+ a b) c) (+ a (+ b c))))
(by-intros a b c)
(by-induction a #:as [() (a-1 IHa)])
;; subgoal 1
reflexivity
;; subgoal 2
display-focus ; show how the context and goal are before rewrite
(by-rewrite IHa)
display-focus ; show how the context and goal are after rewrite
reflexivity)
{{out}}
b : nat
c : nat
a-1 : nat
IHa : (== nat (+ (+ a-1 b) c) (+ a-1 (+ b c)))
--------------------------------
(== nat (S (+ (+ a-1 b) c)) (S (+ a-1 (+ b c))))
b : nat
c : nat
a-1 : nat
IHa : (== nat (+ (+ a-1 b) c) (+ a-1 (+ b c)))
--------------------------------
(== nat (S (+ a-1 (+ b c))) (S (+ a-1 (+ b c))))
The fact that there's no error indicates that all proofs are verified.
Salmon
Note that the only current implementation of Salmon is an interpreter that ignores proofs and doesn't try to check them, but in the future when there is an implementation that checks proofs, it should be able to check the proof in this Salmon code.
pure function even(x) returns boolean ((x in [0...+oo)) && ((x % 2) == 0));
theorem(forall(x : even, y : even) ((x + y) in even))
proof
{
forall (x : even, y : even)
{
L1:
x in even;
L2:
((x in [0...+oo)) && ((x % 2) == 0)) because type_definition(L1);
L3:
((x % 2) == 0) because simplification(L2);
L4:
y in even;
L5:
((y in [0...+oo)) && ((y % 2) == 0)) because type_definition(L4);
L6:
((y % 2) == 0) because simplification(L5);
L7:
(((x + y) % 2) == (x % 2) + (y % 2)); // axiom of % and +
L8:
(((x + y) % 2) == 0 + (y % 2)) because substitution(L3, L7);
L9:
(((x + y) % 2) == 0 + 0) because substitution(L6, L8);
L10:
(((x + y) % 2) == 0) because simplification(L9);
(x + y) in even because type_definition(even, L10);
};
};
Tcl
Using the datatype package from the [[Pattern Matching#Tcl|Pattern Matching]] task...
{{works with|Tcl|8.5}}
package require datatype
datatype define Int = Zero | Succ val
datatype define EO = Even | Odd
proc evenOdd val {
global environment
datatype match $val {
case Zero -> { Even }
case [Succ [Succ x]] -> { evenOdd $x }
case t -> {
set term [list evenOdd $t]
if {[info exists environment($term)]} {
return $environment($term)
} elseif {[info exists environment($t)]} {
return [evenOdd $environment($t)]
} else {
return $term
}
}
}
}
proc add {a b} {
global environment
datatype match $a {
case Zero -> { return $b }
case [Succ x] -> { Succ [add $x $b] }
case t -> {
datatype match $b {
case Zero -> { return $t }
case [Succ x] -> { Succ [add $t $x] }
case t2 -> {
set term [list add $t $t2]
if {[info exists environment($term)]} {
return $environment($term)
} elseif {[info exists environment($t)]} {
return [add $environment($t) $t2]
} elseif {[info exists environment($t2)]} {
return [add $t $environment($t2)]
} else {
return $term
}
}
}
}
}
}
puts "BASE CASE"
puts "evenOdd Zero = [evenOdd Zero]"
puts "evenOdd \[add Zero Zero\] = [evenOdd [add Zero Zero]]"
puts "\nITERATIVE CASE"
set environment([list evenOdd p]) Even
puts "if evenOdd p = Even..."
puts "\tevenOdd \[Succ \[Succ p\]\] = [evenOdd [Succ [Succ p]]]"
unset environment
puts "if evenOdd \[add p q\] = Even..."
set environment([list evenOdd [add p q]]) Even
foreach {a b} {
p q
{Succ {Succ p}} q
p {Succ {Succ q}}
{Succ {Succ p}} {Succ {Succ q}}
} {
puts "\tevenOdd \[[list add $a $b]\] = [evenOdd [add $a $b]]"
}
Output:
BASE CASE
evenOdd Zero = Even
evenOdd [add Zero Zero] = Even
ITERATIVE CASE
if evenOdd p = Even...
evenOdd [Succ [Succ p]] = Even
if evenOdd [add p q] = Even...
evenOdd [add p q] = Even
evenOdd [add {Succ {Succ p}} q] = Even
evenOdd [add p {Succ {Succ q}}] = Even
evenOdd [add {Succ {Succ p}} {Succ {Succ q}}] = Even
It is up to the caller to take the output of this program and interpret it as a proof.
Twelf
nat : type.
z : nat.
s : nat -> nat.
plus : nat -> nat -> nat -> type.
plus-z : plus z N2 N2.
plus-s : plus (s N1) N2 (s N3)
<- plus N1 N2 N3.
%% declare totality assertion
%mode plus +N1 +N2 -N3.
%worlds () (plus _ _ _).
%% check totality assertion
%total N1 (plus N1 _ _).
even : nat -> type.
even-z : even z.
even-s : even (s (s N))
<- even N.
sum-evens : even N1 -> even N2 -> plus N1 N2 N3 -> even N3 -> type.
%mode sum-evens +D1 +D2 +Dplus -D3.
sez : sum-evens
even-z
(DevenN2 : even N2)
(plus-z : plus z N2 N2)
DevenN2.
ses : sum-evens
( (even-s DevenN1') : even (s (s N1')))
(DevenN2 : even N2)
( (plus-s (plus-s Dplus)) : plus (s (s N1')) N2 (s (s N3')))
(even-s DevenN3')
<- sum-evens DevenN1' DevenN2 Dplus DevenN3'.
%worlds () (sum-evens _ _ _ _).
%total D (sum-evens D _ _ _).
{{omit from|Go}} {{omit from|GUISS}} {{omit from|PureBasic}} {{omit from|JavaScript}}