comparison agda/nat.agda @ 77:4b16b485a4b2

Split nat definition
author Yasutaka Higa <e115763@ie.u-ryukyu.ac.jp>
date Mon, 01 Dec 2014 11:58:35 +0900
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children a271f3ff1922
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76:c7076f9bbaed 77:4b16b485a4b2
1 open import Relation.Binary.PropositionalEquality
2 open ≡-Reasoning
3
4 module nat where
5
6 data Nat : Set where
7 O : Nat
8 S : Nat -> Nat
9
10 _+_ : Nat -> Nat -> Nat
11 O + n = n
12 (S m) + n = S (m + n)
13
14 nat-add-right-zero : (n : Nat) -> n ≡ n + O
15 nat-add-right-zero O = refl
16 nat-add-right-zero (S n) = begin
17 S n ≡⟨ cong (\n -> S n) (nat-add-right-zero n) ⟩
18 S (n + O) ≡⟨ refl ⟩
19 S n + O
20
21
22 nat-right-increment : (n m : Nat) -> n + S m ≡ S (n + m)
23 nat-right-increment O m = refl
24 nat-right-increment (S n) m = cong S (nat-right-increment n m)
25
26 nat-add-sym : (n m : Nat) -> n + m ≡ m + n
27 nat-add-sym O O = refl
28 nat-add-sym O (S m) = cong S (nat-add-sym O m)
29 nat-add-sym (S n) O = cong S (nat-add-sym n O)
30 nat-add-sym (S n) (S m) = begin
31 S n + S m ≡⟨ refl ⟩
32 S (n + S m) ≡⟨ cong S (nat-add-sym n (S m)) ⟩
33 S ((S m) + n) ≡⟨ sym (nat-right-increment (S m) n) ⟩
34 S m + S n ∎