Require
Arith.
maximum of two natural numbers |
Fixpoint
max [n:nat] : nat -> nat :=
[m:nat]Cases n m of
O _ => m
| (S n') O => n
| (S n') (S m') => (S (max n' m'))
end.
Simplifications of max
|
Lemma
max_SS : (n,m:nat)((S (max n m))=(max (S n) (S m))).
Proof
.
Auto with arith.
Qed
.
Lemma
max_sym : (n,m:nat)(max n m)=(max m n).
Proof
.
NewInduction n;NewInduction m;Simpl;Auto with arith.
Qed
.
max and le
|
Lemma
max_l : (n,m:nat)(le m n)->(max n m)=n.
Proof
.
NewInduction n;NewInduction m;Simpl;Auto with arith.
Qed
.
Lemma
max_r : (n,m:nat)(le n m)->(max n m)=m.
Proof
.
NewInduction n;NewInduction m;Simpl;Auto with arith.
Qed
.
Lemma
le_max_l : (n,m:nat)(le n (max n m)).
Proof
.
NewInduction n; Intros; Simpl; Auto with arith.
Elim m; Intros; Simpl; Auto with arith.
Qed
.
Lemma
le_max_r : (n,m:nat)(le m (max n m)).
Proof
.
NewInduction n; Simpl; Auto with arith.
NewInduction m; Simpl; Auto with arith.
Qed
.
Hints
Resolve max_r max_l le_max_l le_max_r: arith v62.
max n m is equal to n or m
|
Lemma
max_dec : (n,m:nat){(max n m)=n}+{(max n m)=m}.
Proof
.
NewInduction n;NewInduction m;Simpl;Auto with arith.
Elim (IHn m);Intro H;Elim H;Auto.
Qed
.
Lemma
max_case : (n,m:nat)(P:nat->Set)(P n)->(P m)->(P (max n m)).
Proof
.
NewInduction n; Simpl; Auto with arith.
NewInduction m; Intros; Simpl; Auto with arith.
Pattern (max n m); Apply IHn ; Auto with arith.
Qed
.
Lemma
max_case2 : (n,m:nat)(P:nat->Prop)(P n)->(P m)->(P (max n m)).
Proof
.
NewInduction n; Simpl; Auto with arith.
NewInduction m; Intros; Simpl; Auto with arith.
Pattern (max n m); Apply IHn ; Auto with arith.
Qed
.