Add another example that demonstrates how to use implicit arguments

Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>

examples/ex15.lean

@@ -0,0 +1,47 @@
+
+Variable N : Type
+Variable h : N -> N -> N
+
+Theorem CongrH {a1 a2 b1 b2 : N} (H1 : a1 = b1) (H2 : a2 = b2) : (h a1 a2) = (h b1 b2) :=
+   Congr (Congr (Refl h) H1) H2
+
+(* Display the theorem showing implicit arguments *)
+Set lean::pp::implicit true
+Show Environment 2
+
+(* Display the theorem hiding implicit arguments *)
+Set lean::pp::implicit false
+Show Environment 2
+
+Theorem Example1 (a b c d : N) (H: (a = b ∧ b = c) ∨ (a = d ∧ d = c)) : (h a b) = (h c b) :=
+    DisjCases H
+              (fun H1 : a = b ∧ b = c,
+                   CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
+              (fun H1 : a = d ∧ d = c,
+                   CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
+
+(* Show proof of the last theorem with all implicit arguments *)
+Set lean::pp::implicit true
+Show Environment 1
+
+(* Using placeholders to hide the type of H1 *)
+Theorem Example2 (a b c d : N) (H: (a = b ∧ b = c) ∨ (a = d ∧ d = c)) : (h a b) = (h c b) :=
+    DisjCases H
+              (fun H1 : _,
+                   CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
+              (fun H1 : _,
+                   CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
+
+Set lean::pp::implicit true
+Show Environment 1
+
+(* Same example but the first conjuct has unnecessary stuff *)
+Theorem Example3 (a b c d e : N) (H: (a = b ∧ b = e ∧ b = c) ∨ (a = d ∧ d = c)) : (h a b) = (h c b) :=
+    DisjCases H
+              (fun H1 : _,
+                   CongrH (Trans (Conjunct1 H1) (Conjunct2 (Conjunct2 H1))) (Refl b))
+              (fun H1 : _,
+                   CongrH (Trans (Conjunct1 H1) (Conjunct2 H1)) (Refl b))
+
+Set lean::pp::implicit false
+Show Environment 1