+(* since by assumption H false is equal to (a==b), by rewriting we obtain the goal
+true=(a==b) that is just the boolean version of H1 *)
+
+ <H @\ 5a href="cic:/matita/basics/logic/sym_eq.def(2)"\ 6sym_eq\ 5/a\ 6 @(\b H1)
+
+(* In the "if" case, we proceed by cases over the boolean equality (a==b); if
+(a==b) is false, the goal is trivial; the other case is absurd, since if (a==b) is
+true, then by reflection a=b, while by hypothesis a≠b *)
+
+ |cases (\ 5a href="cic:/matita/basics/bool/true_or_false.def(1)"\ 6true_or_false\ 5/a\ 6 (\ 5a href="cic:/matita/tutorial/chapter4/eqb.fix(0,0,3)"\ 6eqb\ 5/a\ 6 ? a b)) // #H1 @\ 5a href="cic:/matita/basics/logic/False_ind.fix(0,1,1)"\ 6False_ind\ 5/a\ 6 @(\ 5a href="cic:/matita/basics/logic/absurd.def(2)"\ 6absurd\ 5/a\ 6 … (\P H1) H)
+ ]
+qed.
+
+(* We also introduce two operators "\Pf" and "\bf" to reflect a proof
+of (a==b)=false into a proof of a≠b, and vice-versa *)
+
+notation "\Pf H" non associative with precedence 90
+ for @{(proj1 … (eqb_false ???) $H)}.
+
+notation "\bf H" non associative with precedence 90
+ for @{(proj2 … (eqb_false ???) $H)}.
+
+(* The following statement proves that propositional equality in a
+DeqSet is decidable in the traditional sense, namely either a=b or a≠b *)
+
+ \ 5img class="anchor" src="icons/tick.png" id="dec_eq"\ 6lemma dec_eq: ∀S:\ 5a href="cic:/matita/tutorial/chapter4/DeqSet.ind(1,0,0)"\ 6DeqSet\ 5/a\ 6.∀a,b:S. a \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 b \ 5a title="logical or" href="cic:/fakeuri.def(1)"\ 6∨\ 5/a\ 6 a \ 5a title="leibnitz's non-equality" href="cic:/fakeuri.def(1)"\ 6≠\ 5/a\ 6 b.
+#S #a #b cases (\ 5a href="cic:/matita/basics/bool/true_or_false.def(1)"\ 6true_or_false\ 5/a\ 6 (\ 5a href="cic:/matita/tutorial/chapter4/eqb.fix(0,0,3)"\ 6eqb\ 5/a\ 6 ? a b)) #H
+ [%1 @(\P H) | %2 @(\Pf H)]
+qed.
+
+(*
+\ 5h2 class="section"\ 6Unification Hints\ 5/h2\ 6
+A simple example of a set with a decidable equality is bool. We first define
+the boolean equality beqb, that is just the xand function, then prove that
+beqb b1 b2 is true if and only if b1=b2, and finally build the type DeqBool by
+instantiating the DeqSet record with the previous information *)
+
+\ 5img class="anchor" src="icons/tick.png" id="beqb"\ 6definition beqb ≝ λb1,b2.
+ match b1 with [ true ⇒ b2 | false ⇒ \ 5a href="cic:/matita/basics/bool/notb.def(1)"\ 6notb\ 5/a\ 6 b2].
+
+notation < "a == b" non associative with precedence 45 for @{beqb $a $b }.
+
+\ 5img class="anchor" src="icons/tick.png" id="beqb_true"\ 6lemma beqb_true: ∀b1,b2. \ 5a href="cic:/matita/basics/logic/iff.def(1)"\ 6iff\ 5/a\ 6 (\ 5a href="cic:/matita/tutorial/chapter4/beqb.def(2)"\ 6beqb\ 5/a\ 6 b1 b2 \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.con(0,1,0)"\ 6true\ 5/a\ 6) (b1 \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 b2).
+#b1 #b2 cases b1 cases b2 normalize /\ 5span class="autotactic"\ 62\ 5span class="autotrace"\ 6 trace \ 5a href="cic:/matita/basics/logic/And.con(0,1,2)"\ 6conj\ 5/a\ 6\ 5/span\ 6\ 5/span\ 6/
+qed.
+
+\ 5img class="anchor" src="icons/tick.png" id="DeqBool"\ 6definition DeqBool ≝ \ 5a href="cic:/matita/tutorial/chapter4/DeqSet.con(0,1,0)"\ 6mk_DeqSet\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.ind(1,0,0)"\ 6bool\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter4/beqb.def(2)"\ 6beqb\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter4/beqb_true.def(4)"\ 6beqb_true\ 5/a\ 6.
+
+(* At this point, we would expect to be able to prove things like the
+following: for any boolean b, if b==false is true then b=false.
+Unfortunately, this would not work, unless we declare b of type
+DeqBool (change the type in the following statement and see what
+happens). *)
+
+\ 5img class="anchor" src="icons/tick.png" id="exhint"\ 6example exhint: ∀b:\ 5a href="cic:/matita/tutorial/chapter4/DeqBool.def(5)"\ 6DeqBool\ 5/a\ 6. (b\ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a href="cic:/matita/basics/bool/bool.con(0,2,0)"\ 6false\ 5/a\ 6) \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.con(0,1,0)"\ 6true\ 5/a\ 6 → b\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a href="cic:/matita/basics/bool/bool.con(0,2,0)"\ 6false\ 5/a\ 6.
+#b #H @(\P H)
+qed.
+
+(* The point is that == expects in input a pair of objects whose type must be the
+carrier of a DeqSet; bool is indeed the carrier of DeqBool, but the type inference
+system has no knowledge of it (it is an information that has been supplied by the
+user, and stored somewhere in the library). More explicitly, the type inference
+inference system, would face an unification problem consisting to unify bool
+against the carrier of something (a metavaribale) and it has no way to synthetize
+the answer. To solve this kind of situations, matita provides a mechanism to hint
+the system the expected solution. A unification hint is a kind of rule, whose rhd
+is the unification problem, containing some metavariables X1, ..., Xn, and whose
+left hand side is the solution suggested to the system, in the form of equations
+Xi=Mi. The hint is accepted by the system if and only the solution is correct, that
+is, if it is a unifier for the given problem.
+To make an example, in the previous case, the unification problem is bool = carr X,
+and the hint is to take X= mk_DeqSet bool beqb true. The hint is correct, since
+bool is convertible with (carr (mk_DeqSet bool beb true)). *)
+
+unification hint 0 \ 5a href="cic:/fakeuri.def(1)" title="hint_decl_Type1"\ 6≔\ 5/a\ 6 ;
+ X ≟ \ 5a href="cic:/matita/tutorial/chapter4/DeqSet.con(0,1,0)"\ 6mk_DeqSet\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.ind(1,0,0)"\ 6bool\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter4/beqb.def(2)"\ 6beqb\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter4/beqb_true.def(4)"\ 6beqb_true\ 5/a\ 6
+(* ---------------------------------------- *) ⊢
+ \ 5a href="cic:/matita/basics/bool/bool.ind(1,0,0)"\ 6bool\ 5/a\ 6 ≡ \ 5a href="cic:/matita/tutorial/chapter4/carr.fix(0,0,2)"\ 6carr\ 5/a\ 6 X.
+
+unification hint 0 \ 5a href="cic:/fakeuri.def(1)" title="hint_decl_Type0"\ 6≔\ 5/a\ 6 b1,b2:\ 5a href="cic:/matita/basics/bool/bool.ind(1,0,0)"\ 6bool\ 5/a\ 6;
+ X ≟ \ 5a href="cic:/matita/tutorial/chapter4/DeqSet.con(0,1,0)"\ 6mk_DeqSet\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.ind(1,0,0)"\ 6bool\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter4/beqb.def(2)"\ 6beqb\ 5/a\ 6 \ 5a href="cic:/matita/tutorial/chapter4/beqb_true.def(4)"\ 6beqb_true\ 5/a\ 6
+(* ---------------------------------------- *) ⊢
+ \ 5a href="cic:/matita/tutorial/chapter4/beqb.def(2)"\ 6beqb\ 5/a\ 6 b1 b2 ≡ \ 5a href="cic:/matita/tutorial/chapter4/eqb.fix(0,0,3)"\ 6eqb\ 5/a\ 6 X b1 b2.
+
+(* After having provided the previous hints, we may rewrite example exhint
+declaring b of type bool. *)
+
+\ 5img class="anchor" src="icons/tick.png" id="exhint1"\ 6example exhint1: ∀b:\ 5a href="cic:/matita/basics/bool/bool.ind(1,0,0)"\ 6bool\ 5/a\ 6. (b \ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.con(0,2,0)"\ 6false\ 5/a\ 6) \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.con(0,1,0)"\ 6true\ 5/a\ 6 → b \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.con(0,2,0)"\ 6false\ 5/a\ 6.
+#b #H @(\P H)
+qed.
+
+(* The cartesian product of two DeqSets is still a DeqSet. To prove
+this, we must as usual define the boolen equality function, and prove
+it correctly reflects propositional equality. *)
+
+\ 5img class="anchor" src="icons/tick.png" id="eq_pairs"\ 6definition eq_pairs ≝
+ λA,B:\ 5a href="cic:/matita/tutorial/chapter4/DeqSet.ind(1,0,0)"\ 6DeqSet\ 5/a\ 6.λp1,p2:A\ 5a title="Product" href="cic:/fakeuri.def(1)"\ 6×\ 5/a\ 6B.(\ 5a title="pair pi1" href="cic:/fakeuri.def(1)"\ 6\fst\ 5/a\ 6 p1 \ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a title="pair pi1" href="cic:/fakeuri.def(1)"\ 6\fst\ 5/a\ 6 p2) \ 5a title="boolean and" href="cic:/fakeuri.def(1)"\ 6∧\ 5/a\ 6 (\ 5a title="pair pi2" href="cic:/fakeuri.def(1)"\ 6\snd\ 5/a\ 6 p1 \ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a title="pair pi2" href="cic:/fakeuri.def(1)"\ 6\snd\ 5/a\ 6 p2).
+
+\ 5img class="anchor" src="icons/tick.png" id="eq_pairs_true"\ 6lemma eq_pairs_true: ∀A,B:\ 5a href="cic:/matita/tutorial/chapter4/DeqSet.ind(1,0,0)"\ 6DeqSet\ 5/a\ 6.∀p1,p2:A\ 5a title="Product" href="cic:/fakeuri.def(1)"\ 6×\ 5/a\ 6B.
+ \ 5a href="cic:/matita/tutorial/chapter4/eq_pairs.def(4)"\ 6eq_pairs\ 5/a\ 6 A B p1 p2 \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 \ 5a href="cic:/matita/basics/bool/bool.con(0,1,0)"\ 6true\ 5/a\ 6 \ 5a title="iff" href="cic:/fakeuri.def(1)"\ 6↔\ 5/a\ 6 p1 \ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6 p2.
+#A #B * #a1 #b1 * #a2 #b2 %
+ [#H cases (\ 5a href="cic:/matita/basics/bool/andb_true.def(5)"\ 6andb_true\ 5/a\ 6 …H) normalize #eqa #eqb >(\P eqa) >(\P eqb) //
+ |#H destruct normalize >(\b (\ 5a href="cic:/matita/basics/logic/eq.con(0,1,2)"\ 6refl\ 5/a\ 6 … a2)) >(\b (\ 5a href="cic:/matita/basics/logic/eq.con(0,1,2)"\ 6refl\ 5/a\ 6 … b2)) //
+ ]
+qed.
+
+\ 5img class="anchor" src="icons/tick.png" id="DeqProd"\ 6definition DeqProd ≝ λA,B:\ 5a href="cic:/matita/tutorial/chapter4/DeqSet.ind(1,0,0)"\ 6DeqSet\ 5/a\ 6.
+ \ 5a href="cic:/matita/tutorial/chapter4/DeqSet.con(0,1,0)"\ 6mk_DeqSet\ 5/a\ 6 (A\ 5a title="Product" href="cic:/fakeuri.def(1)"\ 6×\ 5/a\ 6B) (\ 5a href="cic:/matita/tutorial/chapter4/eq_pairs.def(4)"\ 6eq_pairs\ 5/a\ 6 A B) (\ 5a href="cic:/matita/tutorial/chapter4/eq_pairs_true.def(6)"\ 6eq_pairs_true\ 5/a\ 6 A B).
+
+(* Having an unification problem of the kind T1×T2 = carr X, what kind
+of hint can we give to the system? We expect T1 to be the carrier of a
+DeqSet C1, T2 to be the carrier of a DeqSet C2, and X to be DeqProd C1 C2.
+This is expressed by the following hint: *)
+
+unification hint 0 \ 5a href="cic:/fakeuri.def(1)" title="hint_decl_Type1"\ 6≔\ 5/a\ 6 C1,C2;
+ T1 ≟ \ 5a href="cic:/matita/tutorial/chapter4/carr.fix(0,0,2)"\ 6carr\ 5/a\ 6 C1,
+ T2 ≟ \ 5a href="cic:/matita/tutorial/chapter4/carr.fix(0,0,2)"\ 6carr\ 5/a\ 6 C2,
+ X ≟ \ 5a href="cic:/matita/tutorial/chapter4/DeqProd.def(7)"\ 6DeqProd\ 5/a\ 6 C1 C2
+(* ---------------------------------------- *) ⊢
+ T1\ 5a title="Product" href="cic:/fakeuri.def(1)"\ 6×\ 5/a\ 6T2 ≡ \ 5a href="cic:/matita/tutorial/chapter4/carr.fix(0,0,2)"\ 6carr\ 5/a\ 6 X.
+
+unification hint 0 \ 5a href="cic:/fakeuri.def(1)" title="hint_decl_Type0"\ 6≔\ 5/a\ 6 T1,T2,p1,p2;
+ X ≟ \ 5a href="cic:/matita/tutorial/chapter4/DeqProd.def(7)"\ 6DeqProd\ 5/a\ 6 T1 T2
+(* ---------------------------------------- *) ⊢
+ \ 5a href="cic:/matita/tutorial/chapter4/eq_pairs.def(4)"\ 6eq_pairs\ 5/a\ 6 T1 T2 p1 p2 ≡ \ 5a href="cic:/matita/tutorial/chapter4/eqb.fix(0,0,3)"\ 6eqb\ 5/a\ 6 X p1 p2.
+
+\ 5img class="anchor" src="icons/tick.png" id="hint2"\ 6example hint2: ∀b1,b2.
+ \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"\ 6〈\ 5/a\ 6b1,\ 5a href="cic:/matita/basics/bool/bool.con(0,1,0)"\ 6true\ 5/a\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"\ 6〉\ 5/a\ 6\ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a title="eqb" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"\ 6〈\ 5/a\ 6\ 5a href="cic:/matita/basics/bool/bool.con(0,2,0)"\ 6false\ 5/a\ 6,b2\ 5a title="Pair construction" 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/basics/bool/bool.con(0,1,0)"\ 6true\ 5/a\ 6 → \ 5a title="Pair construction" href="cic:/fakeuri.def(1)"\ 6〈\ 5/a\ 6b1,\ 5a href="cic:/matita/basics/bool/bool.con(0,1,0)"\ 6true\ 5/a\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"\ 6〉\ 5/a\ 6\ 5a title="leibnitz's equality" href="cic:/fakeuri.def(1)"\ 6=\ 5/a\ 6\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"\ 6〈\ 5/a\ 6\ 5a href="cic:/matita/basics/bool/bool.con(0,2,0)"\ 6false\ 5/a\ 6,b2\ 5a title="Pair construction" href="cic:/fakeuri.def(1)"\ 6〉\ 5/a\ 6.
+#b1 #b2 #H @(\P H).
\ No newline at end of file