Proof

Define a type for natural numbers (0, 1, 2, 3, ...) representable by a computer, and addition on them. Define a type of even numbers (0, 2, 4, 6, ...) within the previously defined range of natural numbers. Prove that the addition of any two even numbers is even, if the result is a member of the type.

Agda

Requires version w.

<lang agda>module Arith where

data Nat : Set where

 zero : Nat
 suc  : Nat -> Nat

_+_ : Nat -> Nat -> Nat zero + n = n suc m + n = suc (m + n)

data Even : Nat -> Set where

 even_zero    : Even zero
 even_suc_suc : {n : Nat} -> Even n -> Even (suc (suc n))

_even+_ : {m n : Nat} -> Even m -> Even n -> Even (m + n) even_zero even+ en = en even_suc_suc em even+ en = even_suc_suc (em even+ en) </lang>

Coq

<lang coq>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. </lang>

Haskell

See Proof/Haskell.

Omega

<lang 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) </lang>

Tcl

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

Works with: Tcl version 8.5

<lang tcl>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]]"

}</lang> 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

<lang 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 _ _ _). </lang>