Proof

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Proof is a draft programming task. It is not yet considered ready to be promoted as a complete task, for reasons that should be found in its talk page.

This task only makes sense for dependently-typed languages and 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 Peano axioms using the language itself, that is a way to construct natural and even natural numbers in the 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 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 constructive proofs. Examples of the trusted mathematical metatheories can be given: SystemF for Haskell, MLTT for Agda, CoC/CoIC for Coq.

Task:

  1. To illustrate the possibility of type formation and type introduction:
    1. Define a countably infinite set of natural numbers {0, 1, 2, 3, ...}.
    2. Define a countably infinite set of even natural numbers {0, 2, 4, 6, ...} within the previously defined set of natural numbers.
    3. Define a countably infinite set of odd natural numbers {1, 3, 5, 7, ...} within the previously defined set of natural numbers.
  2. To illustrate the possibility of type elimination:
    1. Define the addition on natural numbers.
  3. To demonstrate constructive proofs:
    1. Prove that the addition of any two even numbers is even.
    2. Prove that the addition is always associative.
    3. Prove that the addition is always commutative.
    4. Try to prove that the addition of any two even numbers is odd (it should be rejected).
  4. To demonstrate the ability of disproofs:
    1. Prove that the addition of any two even numbers cannot be odd.
    2. Try to prove that the addition of any two even numbers cannot be even (it should be rejected).

3. and 4. can't be done using a simple number enumeration since there is a countable many natural numbers which is quantified in propositions.

Contents

[edit] ACL2

3.1. using built-in natural numbers:

(thm (implies (and (evenp x) (evenp y))
(evenp (+ x y))))

[edit] 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)
--

[edit] 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.
 

[edit] Haskell

Using GADTs and 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.

[edit] 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

  1. all natural numbers are even, and
  2. addition is associative, and
  3. addition is commutative, and
  4. 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.

[edit] Mathematica

 
(*1.1*)
natural[0] := True;
natural[s[x_]] := natural[x];
 
(*1.2*)
even[0] := True;
even[s[0]] := False;
even[s[s[x_]]] := even[x];
 
(*1.3*)
odd[x_] := ! even[x];
 
(*2.1*)
plus[x_, 0] := x;
plus[x_, s[y_]] := s[plus[x, y]];
 
(*Induction*)
proof[statement_, x_,
s_] := (statement /. x -> 0) &&
Implies[statement,
statement /. x -> s[x]] //. {Implies[t1_ == t2_,
f_[t1_] == f_[t2_]] :> True}
 
(*3.1*)
proof[even[x] \[Implies] even[plus[x, y]], y, s[s[#]] &]
(*Output: True*)
 
(*3.2*)
proof[plus[x, plus[y, z]] == plus[plus[x, y], z], z, s]
(*Output: True*)
 
(*3.3*)
 
 
(*3.4*)
proof[even[x] \[Implies] odd[plus[x, y]], y, s[s[#]] &]
(*Output: even[x]\[Implies]!even[x]*)
 
(*4.1*)
proof[even[x] \[Implies] ! odd[plus[x, y]], y, s[s[#]] &]
(*Output: True*)
 
(*4.2*)
proof[even[x] \[Implies] ! even[plus[x, y]], y, s[s[#]] &]
(*Output: even[x]\[Implies]!even[x]*)
 

[edit] 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)
 

[edit] 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);
};
};

[edit] Tcl

Using the datatype package from the Pattern Matching task...

Works with: Tcl version 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.

[edit] 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 _ _ _).
 
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