1
(************************************************************************)
2
(* v * The Coq Proof Assistant / The Coq Development Team *)
3
(* <O___,, * CNRS-Ecole Polytechnique-INRIA Futurs-Universite Paris Sud *)
4
(* \VV/ **************************************************************)
5
(* // * This file is distributed under the terms of the *)
6
(* * GNU Lesser General Public License Version 2.1 *)
7
(************************************************************************)
8
(* Evgeny Makarov, INRIA, 2007 *)
9
(************************************************************************)
11
(*i $Id: NAdd.v 11674 2008-12-12 19:48:40Z letouzey $ i*)
15
Module NAddPropFunct (Import NAxiomsMod : NAxiomsSig).
16
Module Export NBasePropMod := NBasePropFunct NAxiomsMod.
18
Open Local Scope NatScope.
21
forall n1 n2 : N, n1 == n2 -> forall m1 m2 : N, m1 == m2 -> n1 + m1 == n2 + m2.
24
Theorem add_0_l : forall n : N, 0 + n == n.
27
Theorem add_succ_l : forall n m : N, (S n) + m == S (n + m).
30
(** Theorems that are valid for both natural numbers and integers *)
32
Theorem add_0_r : forall n : N, n + 0 == n.
35
Theorem add_succ_r : forall n m : N, n + S m == S (n + m).
38
Theorem add_comm : forall n m : N, n + m == m + n.
41
Theorem add_assoc : forall n m p : N, n + (m + p) == (n + m) + p.
44
Theorem add_shuffle1 : forall n m p q : N, (n + m) + (p + q) == (n + p) + (m + q).
47
Theorem add_shuffle2 : forall n m p q : N, (n + m) + (p + q) == (n + q) + (m + p).
50
Theorem add_1_l : forall n : N, 1 + n == S n.
53
Theorem add_1_r : forall n : N, n + 1 == S n.
56
Theorem add_cancel_l : forall n m p : N, p + n == p + m <-> n == m.
59
Theorem add_cancel_r : forall n m p : N, n + p == m + p <-> n == m.
62
(* Theorems that are valid for natural numbers but cannot be proved for Z *)
64
Theorem eq_add_0 : forall n m : N, n + m == 0 <-> n == 0 /\ m == 0.
67
(* The next command does not work with the axiom add_0_l from NAddSig *)
68
rewrite add_0_l. intuition reflexivity.
69
intros n IH. rewrite add_succ_l.
70
setoid_replace (S (n + m) == 0) with False using relation iff by
71
(apply -> neg_false; apply neq_succ_0).
72
setoid_replace (S n == 0) with False using relation iff by
73
(apply -> neg_false; apply neq_succ_0). tauto.
77
forall n m : N, (exists p : N, n + m == S p) <->
78
(exists n' : N, n == S n') \/ (exists m' : N, m == S m').
82
destruct H as [p H]. rewrite add_0_l in H; right; now exists p.
83
destruct H as [[n' H] | [m' H]].
84
symmetry in H; false_hyp H neq_succ_0.
85
exists m'; now rewrite add_0_l.
86
intro n; split; intro H.
88
exists (n + m); now rewrite add_succ_l.
91
Theorem eq_add_1 : forall n m : N,
92
n + m == 1 -> n == 1 /\ m == 0 \/ n == 0 /\ m == 1.
95
assert (H1 : exists p : N, n + m == S p) by now exists 0.
96
apply -> eq_add_succ in H1. destruct H1 as [[n' H1] | [m' H1]].
97
left. rewrite H1 in H; rewrite add_succ_l in H; apply succ_inj in H.
98
apply -> eq_add_0 in H. destruct H as [H2 H3]; rewrite H2 in H1; now split.
99
right. rewrite H1 in H; rewrite add_succ_r in H; apply succ_inj in H.
100
apply -> eq_add_0 in H. destruct H as [H2 H3]; rewrite H3 in H1; now split.
103
Theorem succ_add_discr : forall n m : N, m ~= S (n + m).
106
apply neq_sym. apply neq_succ_0.
107
intros m IH H. apply succ_inj in H. rewrite add_succ_r in H.
108
unfold not in IH; now apply IH.
111
Theorem add_pred_l : forall n m : N, n ~= 0 -> P n + m == P (n + m).
115
intros n IH; rewrite add_succ_l; now do 2 rewrite pred_succ.
118
Theorem add_pred_r : forall n m : N, m ~= 0 -> n + P m == P (n + m).
120
intros n m H; rewrite (add_comm n (P m));
121
rewrite (add_comm n m); now apply add_pred_l.
124
(* One could define n <= m as exists p : N, p + n == m. Then we have
127
forall n m : N, n <= m \/ m <= n,
131
forall n m : N, (exists p : N, p + n == m) \/ (exists p : N, p + m == n) (1)
133
We will need (1) in the proof of induction principle for integers
134
constructed as pairs of natural numbers. The formula (1) can be proved
135
using properties of order and truncated subtraction. Thus, p would be
136
m - n or n - m and (1) would hold by theorem sub_add from Sub.v
137
depending on whether n <= m or m <= n. However, in proving induction
138
for integers constructed from natural numbers we do not need to
139
require implementations of order and sub; it is enough to prove (1)
142
Theorem add_dichotomy :
143
forall n m : N, (exists p : N, p + n == m) \/ (exists p : N, p + m == n).
145
intros n m; induct n.
146
left; exists m; apply add_0_r.
148
destruct IH as [[p H] | [p H]].
149
destruct (zero_or_succ p) as [H1 | [p' H1]]; rewrite H1 in H.
150
rewrite add_0_l in H. right; exists (S 0); rewrite H; rewrite add_succ_l; now rewrite add_0_l.
151
left; exists p'; rewrite add_succ_r; now rewrite add_succ_l in H.
152
right; exists (S p). rewrite add_succ_l; now rewrite H.