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Revision as of 00:33, 9 September 2009
You are encouraged to solve this task according to the task description, using any language you may know.
Define a type for a sub-set of 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.
Agda2
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)
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.
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)
Tcl
Using the datatype
package from the Pattern Matching task...
<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
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 _ _ _).