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
+The elim tactic, allow you to apply induction in a very 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).
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. *)
+(* 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).
#n elim n normalize