+let rec length (A:Type[0]) (l:\ 5a href="cic:/matita/tutorial/chapter3/list.ind(1,0,1)"\ 6list\ 5/a\ 6 A) on l ≝
+match l with
+ [ nil ⇒ \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"\ 6O\ 5/a\ 6
+ | cons a tl ⇒ \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"\ 6S\ 5/a\ 6 (length A tl)].
+
+let rec nth n (A:Type[0]) (l:\ 5a href="cic:/matita/tutorial/chapter3/list.ind(1,0,1)"\ 6list\ 5/a\ 6 A) (d:A) ≝
+ match n with
+ [O ⇒ \ 5a href="cic:/matita/tutorial/chapter3/head.def(1)"\ 6head\ 5/a\ 6 A l d
+ |S m ⇒ nth m A (\ 5a href="cic:/matita/tutorial/chapter3/tail.def(1)"\ 6tail\ 5/a\ 6 A l) d].
+
+example ex_length: \ 5a href="cic:/matita/tutorial/chapter3/length.fix(0,1,1)"\ 6length\ 5/a\ 6 ? (\ 5a href="cic:/matita/tutorial/chapter3/list.con(0,2,1)"\ 6cons\ 5/a\ 6 ? \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"\ 6O\ 5/a\ 6 \ 5a title="nil" href="cic:/fakeuri.def(1)"\ 6[\ 5/a\ 6]) \ 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/nat.con(0,1,0)"\ 6O\ 5/a\ 6.
+normalize // qed.
+
+example ex_nth: \ 5a href="cic:/matita/tutorial/chapter3/nth.fix(0,0,2)"\ 6nth\ 5/a\ 6 (\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"\ 6S\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"\ 6O\ 5/a\ 6) ? (\ 5a href="cic:/matita/tutorial/chapter3/list.con(0,2,1)"\ 6cons\ 5/a\ 6 ? (\ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,2,0)"\ 6S\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"\ 6O\ 5/a\ 6) (\ 5a href="cic:/matita/tutorial/chapter3/list.con(0,2,1)"\ 6cons\ 5/a\ 6 ? \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"\ 6O\ 5/a\ 6 \ 5a title="nil" href="cic:/fakeuri.def(1)"\ 6[\ 5/a\ 6])) \ 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 \ 5a href="cic:/matita/tutorial/chapter2/nat.con(0,1,0)"\ 6O\ 5/a\ 6.
+normalize // qed.
+
+(* Proving that the length of l1@l2 is the sum of the lengths of l1
+and l2 just requires a trivial induction on the first list. *)
+
+ lemma length_add: ∀A.∀l1,l2:\ 5a href="cic:/matita/tutorial/chapter3/list.ind(1,0,1)"\ 6list\ 5/a\ 6 A.
+ \ 5a href="cic:/matita/tutorial/chapter3/length.fix(0,1,1)"\ 6length\ 5/a\ 6 ? (l1\ 5a title="append" href="cic:/fakeuri.def(1)"\ 6@\ 5/a\ 6l2) \ 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 (\ 5a href="cic:/matita/tutorial/chapter3/length.fix(0,1,1)"\ 6length\ 5/a\ 6 ? l1) (\ 5a href="cic:/matita/tutorial/chapter3/length.fix(0,1,1)"\ 6length\ 5/a\ 6 ? l2).
+#A #l1 elim l1 normalize // qed.
+
+(* Let us come to a more interesting question. How can we prove that the empty list is
+different from any list with at least one element, that is from any list of the kind (a::l)?
+We start defining a simple predicate stating if a list is empty or not. The predicate
+is computed by inspection over the list *)
+
+definition is_nil: ∀A:Type[0].\ 5a href="cic:/matita/tutorial/chapter3/list.ind(1,0,1)"\ 6list\ 5/a\ 6 A → Prop ≝
+λA.λl.match l with [ nil ⇒ l \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a title="nil" href="cic:/fakeuri.def(1)"\ 6[\ 5/a\ 6] | cons hd tl ⇒ (l \ 5a title="leibnitz's non-equality" href="cic:/fakeuri.def(1)"\ 6≠\ 5/a\ 6 \ 5a title="nil" href="cic:/fakeuri.def(1)"\ 6[\ 5/a\ 6])].
+
+(* Next we need a simple result about negation: if you wish to prove ¬P you are
+authorized to add P to your hypothesis: *)
+
+lemma neg_aux : ∀P:Prop. (P → \ 5a title="logical not" href="cic:/fakeuri.def(1)"\ 6¬\ 5/a\ 6P) → \ 5a title="logical not" href="cic:/fakeuri.def(1)"\ 6¬\ 5/a\ 6P.
+#P #PtonegP % /\ 5span class="autotactic"\ 63\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/basics/logic/absurd.def(2)"\ 6absurd\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/ qed.
+
+theorem diff_cons_nil:
+∀A:Type[0].∀l:\ 5a href="cic:/matita/tutorial/chapter3/list.ind(1,0,1)"\ 6list\ 5/a\ 6 A.∀a:A. a\ 5a title="cons" href="cic:/fakeuri.def(1)"\ 6:\ 5/a\ 6:l \ 5a title="leibnitz's non-equality" href="cic:/fakeuri.def(1)"\ 6≠\ 5/a\ 6 \ 5a title="nil" href="cic:/fakeuri.def(1)"\ 6[\ 5/a\ 6].
+#A #l #a @\ 5a href="cic:/matita/tutorial/chapter3/neg_aux.def(3)"\ 6neg_aux\ 5/a\ 6 #Heq
+(* we start assuming the new hypothesis Heq of type a::l = [] using neg_aux.
+Next we use the change tactic to pass from the current goal a::l≠ [] to the expression
+is_nil a::l, convertible with it. *)
+(change with (\ 5a href="cic:/matita/tutorial/chapter3/is_nil.def(1)"\ 6is_nil\ 5/a\ 6 ? (a\ 5a title="cons" href="cic:/fakeuri.def(1)"\ 6:\ 5/a\ 6:l)))
+(* Now, we rewrite with Heq, obtaining (is_nil A []), that reduces to the trivial
+goal [] = [] *)
+>Heq // qed.
+
+(* As an application of the previous result let us prove that l1@l2 is empty if and
+only if both l1 and l2 are empty. The idea is to proceed by cases on l1: if l1=[] the
+statement is trivial; on the other side, if l1 = a::tl, then the hypothesis
+(a::tl)@l2 = [] is absurd, hence we can prove anything from it. When we know we can
+prove both A and ¬A, a sensible way to proceed is to apply False_ind: ∀P.False → P to the
+current goal, that breaks down to prove False, and then absurd: ∀A:Prop. A → ¬A → False
+to reduce to the contradictory cases. Usually, you may invoke automation to take care
+to solve the absurd case. *)
+
+lemma nil_to_nil: ∀A.∀l1,l2:\ 5a href="cic:/matita/tutorial/chapter3/list.ind(1,0,1)"\ 6list\ 5/a\ 6 \ 5span style="text-decoration: underline;"\ 6\ 5/span\ 6A.