-(*
- ||M|| This file is part of HELM, an Hypertextual, Electronic
- ||A|| Library of Mathematics, developed at the Computer Science
- ||T|| Department of the University of Bologna, Italy.
- ||I||
- ||T||
- ||A|| This file is distributed under the terms of the
- \ / GNU General Public License Version 2
- \ /
- V_______________________________________________________________ *)
-
-include "arithmetics/nat.ma".
+include "basics/logic.ma".
+
+(* Most of the types we have seen so far are enumerated types, composed by
+a finite set of alternatives, and records, composed by tuples of
+heteregoneous elements. A more interesting case of type definition is
+when some of the rules defining its elements are recursive, i.e. they
+allow the formation of more elements of the type in terms of the already
+defined ones. The most typical case is provided by the natural numbers,
+that can be defined as the smallest set generated by a constant 0 and a
+successor function from natural numbers to natural numbers *)
+
+inductive nat : Type[0] ≝
+| O :nat
+| S: nat →nat.
+
+(* The two terms O and S are called constructors: they define the
+signature of the type, whose objects are the elements freely generated
+by means of them. So, examples of natural numbers are O, S O, S (S O),
+S (S (S O)) and so on.
+
+The language of Matita allows the definition of well founded recursive functions
+over inductive types; in order to guarantee termination of recursion you are
+only allowed to make recursive calls on structurally smaller arguments than the
+ones your received in input.
+Most mathematical functions can be naturally defined in this way. For instance,
+the sum of two natural numbers can be defined as follows *)
+
+let rec add n m ≝
+match n with
+[ O ⇒ m
+| S a ⇒ \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"\ 6S\ 5/a\ 6 (add a m)
+].
+
+(* It is worth to observe that the previous algorithm works by recursion over the
+first argument. This means that, for instance, (add 0 x) will reduce to x, as expected,
+but (add x 0) is stack. How can we prove that, for a generic x, (add x 0) = x? The
+mathematical tool do it is called induction. The induction principle states that,
+given a property P(n) over natural numbers, if we prove P(0) and prove that, for any
+m, P(m) implies P(S m), than we can conclude P(n) for any n.
+
+The elim tactic, allow you to apply induction in a vcery simple way. If your
+goal is P(n), the invocation of
+ elim n
+will break down your task to prove the two subgoals P(0) and ∀m.P(m) → P(S m).
+Let us apply it to our case *)
+
+lemma add_0: ∀a. \ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"\ 6add\ 5/a\ 6 a \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"\ 6O\ 5/a\ 6 \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 a.
+#a elim a
+
+(* If you stop the computation here, you will see on the right the two subgoals
+ - add O O = O
+ - ∀x. add x 0 = x → add (S x) O = S x
+After normalization, both goals are trivial.
+*)
+
+normalize // qed.
+
+ (* In a similar way, it is convenient to state a lemma about the behaviour of add when the
+second argument is not zero. *)
+
+lemma add_S : ∀a,b. \ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"\ 6add\ 5/a\ 6 a (\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"\ 6S\ 5/a\ 6 b) \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"\ 6S\ 5/a\ 6 (\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"\ 6add\ 5/a\ 6 a b).
+
+(* In the same way as before, we proceed by induction over a. *)
+
+#a #b elim a normalize //
+qed.
+
+(* We are now in the position to prove the commutativity of the sum *)
+
+theorem add_comm : ∀a,b. \ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"\ 6add\ 5/a\ 6 a b \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"\ 6add\ 5/a\ 6 b a.
+#a elim a normalize
+
+(* We have two sub goals:
+ G1: ∀b. b = add b O
+ G2: ∀x.(∀b. add x b = add b x) → ∀b. S (add x b) = add b (S x).
+G1 is just our lemma add_O. For G2, we start introducing x and the induction
+hypothesis IH; then, the goal is proved by rewriting using add_S and IH.
+For Matita, the task is trivial and we can simply close the goal with // *)
+
+// qed.
+
+(* Let us now define the following function: *)
+
+definition twice ≝ λn.\ 5a href="cic:/matita/tutorial/chapter2/add.fix(0,0,1)"\ 6add\ 5/a\ 6 n n.
+
+(* We are interested to prove that for any natural number m there
+exists a natural number m that is the integer half of m. This
+will give us the opportunity to introduce new connectives
+and quantifiers, and later on to make some interesting consideration
+on proofs and computations. *)
+
+theorem ex_half: ∀n.\ 5a title="exists" href="cic:/fakeuri.def(1)"\ 6∃\ 5/a\ 6m. n \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"\ 6twice\ 5/a\ 6 m \ 5a title="logical or" href="cic:/fakeuri.def(1)"\ 6∨\ 5/a\ 6 n \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"\ 6S\ 5/a\ 6 (\ 5a href="cic:/matita/tutorial/chapter2/twice.def(2)"\ 6twice\ 5/a\ 6 m).
+
+
+
+ include "basics.ma".
include "basics/list.ma".
interpretation "iff" 'iff a b = (iff a b).