(* *)
(**************************************************************************)
-set "baseuri" "cic:/matita/test/decl".
+
include "nat/times.ma".
include "nat/orders.ma".
]
qed.
+theorem easy15: ∀n,m. n * m = O → n = O ∨ m = O.
+ assume n: nat.
+ assume m: nat.
+ (* base case *)
+ by _ we proved (O = O) (trivial).
+ by _ we proved (O = O ∨ m = O) (trivial2).
+ by _ we proved (O*m=O → O=O ∨ m=O) (base_case).
+ (* inductive case *)
+ we need to prove
+ (∀n1. (n1 * m = O → n1 = O ∨ m = O) → (S n1) * m = O → (S n1) = O ∨ m = O)
+ (inductive_case).
+ assume n1: nat.
+ suppose (n1 * m = O → n1 = O ∨ m = O) (inductive_hyp).
+ (* base case *)
+ by _ we proved (S n1 = O ∨ O = O) (pre_base_case2).
+ by _ we proved (S n1*O = O → S n1 = O ∨ O = O) (base_case2).
+ (* inductive case *)
+ we need to prove
+ (∀m1. (S n1 * m1 = O → S n1 = O ∨ m1 = O) →
+ (S n1 * S m1 = O → S n1 = O ∨ S m1 = O)) (inductive_hyp2).
+ assume m1: nat.
+ suppose (S n1 * m1 = O → S n1 = O ∨ m1 = O) (useless).
+ suppose (S n1 * S m1 = O) (absurd_hyp).
+ simplify in absurd_hyp.
+ by _ we proved (O = S (m1+n1*S m1)) (absurd_hyp').
+ (* BUG: automation failure *)
+ by (not_eq_O_S ? absurd_hyp') we proved False (the_absurd).
+ (* BUG: automation failure *)
+ by (False_ind ? the_absurd)
+ done.
+ (* the induction *)
+ by (nat_ind (λm.S n1 * m = O → S n1 = O ∨ m = O) base_case2 inductive_hyp2 m)
+ done.
+ (* the induction *)
+ by (nat_ind (λn.n*m=O → n=O ∨ m=O) base_case inductive_case n)
+done.
+qed.
theorem easy3: ∀A:Prop. (A ∧ ∃n:nat.n ≠ n) → True.
assume P: Prop.
suppose (P ∧ ∃m:nat.m ≠ m) (H).
by H we have P (H1) and (∃x:nat.x≠x) (H2).
- (*BUG:
by H2 let q:nat such that (q ≠ q) (Ineq).
- *)
- (* the next line is wrong, but for the moment it does the job *)
- by H2 let q:nat such that False (Ineq).
by I done.
qed.
suppose (n=m) (H).
suppose (S m = S p) (K).
suppose (n = S p) (L).
-obtain (S n) = (S m) by (eq_f ? ? ? ? ? H).
+conclude (S n) = (S m) by H.
= (S p) by K.
- = n by (sym_eq ? ? ? L)
+ = n by L
+done.
+qed.
+
+theorem easy45: ∀n,m,p. n = m → S m = S p → n = S p → S n = n.
+assume n: nat.
+assume m:nat.
+assume p:nat.
+suppose (n=m) (H).
+suppose (S m = S p) (K).
+suppose (n = S p) (L).
+conclude (S n) = (S m).
+ = (S p).
+ = n
done.
qed.
+
+theorem easy5: ∀n:nat. n*O=O.
+assume n: nat.
+(* Bug here: False should be n*0=0 *)
+we proceed by induction on n to prove False.
+ case O.
+ the thesis becomes (O*O=O).
+ by (refl_eq ? O) done.
+ case S (m:nat).
+ by induction hypothesis we know (m*O=O) (I).
+ the thesis becomes (S m * O = O).
+ (* Bug here: missing that is equivalent to *)
+ simplify.
+ by I done.
+qed.
+
+inductive tree : Type ≝
+ Empty: tree
+ | Node: tree → tree → tree.
+
+let rec size t ≝
+ match t with
+ [ Empty ⇒ O
+ | (Node t1 t2) ⇒ S ((size t1) + (size t2))
+ ].
+
+theorem easy6: ∀t. O ≮ O → O < size t → t ≠ Empty.
+ assume t: tree.
+ suppose (O ≮ O) (trivial).
+ (*Bug here: False should be something else *)
+ we proceed by induction on t to prove False.
+ case Empty.
+ the thesis becomes (O < size Empty → Empty ≠ Empty).
+ suppose (O < size Empty) (absurd)
+ that is equivalent to (O < O).
+ (* Here the "natural" language is not natural at all *)
+ we proceed by induction on (trivial absurd) to prove False.
+ (*Bug here: this is what we want
+ case Node (t1:tree) (t2:tree).
+ by induction hypothesis we know (O < size t1 → t1 ≠ Empty) (Ht1).
+ by induction hypothesis we know (O < size t2 → t2 ≠ Empty) (Ht2). *)
+ (*This is the best we can do right now*)
+ case Node.
+ assume t1: tree.
+ by induction hypothesis we know (O < size t1 → t1 ≠ Empty) (Ht1).
+ assume t2: tree.
+ by induction hypothesis we know (O < size t2 → t2 ≠ Empty) (Ht2).
+ suppose (O < size (Node t1 t2)) (useless).
+ we need to prove (Node t1 t2 ≠ Empty) (final)
+ or equivalently (Node t1 t2 = Empty → False).
+ suppose (Node t1 t2 = Empty) (absurd).
+ (* Discriminate should really generate a theorem to be useful with
+ declarative tactics *)
+ destruct absurd.
+ by final
+ done.
+qed.
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