feat(library/elaborator): only expand definitions that are not marked as hidden

The elaborator produces better proof terms. This is particularly important when we have to prove the remaining holes using tactics.
For example, in one of the tests, the elaborator was producing the sub-expression

 (λ x : N, if ((λ x::1 : N, if (P a x x::1) ⊥ ⊤) == (λ x : N, ⊤)) ⊥ ⊤)

After, this commit it produces

 (λ x : N, ¬ ∀ x::1 : N, ¬ P a x x::1)

The expressions above are definitionally equal, but the second is easier to work with.

Question: do we really need hidden definitions?
Perhaps, we can use only the opaque flag.

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

src/library/elaborator/elaborator.cpp

@@ -19,6 +19,7 @@ Author: Leonardo de Moura
 #include "kernel/replace_fn.h"
 #include "kernel/builtin.h"
 #include "kernel/update_expr.h"
+#include "library/hidden_defs.h"
 #include "library/type_inferer.h"
 #include "library/elaborator/elaborator.h"
 #include "library/elaborator/elaborator_justification.h"
@@ -626,7 +627,7 @@ class elaborator::imp {
     int get_const_weight(expr const & a) {
         lean_assert(is_constant(a));
         optional<object> obj = m_env->find_object(const_name(a));
-        if (obj && obj->is_definition() && !obj->is_opaque())
+        if (obj && obj->is_definition() && !obj->is_opaque() && !is_hidden(m_env, const_name(a)))
             return obj->get_weight();
         else
             return -1;

src/library/hidden_defs.cpp

@@ -63,7 +63,7 @@ void set_hidden_flag(environment const & env, name const & d, bool flag) {
 
 void hide_builtin(environment const & env) {
     for (auto c : { mk_implies_fn(), mk_iff_fn(), mk_not_fn(), mk_or_fn(), mk_and_fn(),
-                mk_forall_fn(), mk_exists_fn(), mk_homo_eq_fn() })
+                mk_forall_fn() })
         set_hidden_flag(env, const_name(c));
 }
 

tests/lean/elab1.lean.expected.out

@@ -130,12 +130,8 @@ A : Type, B : Type, a : ?M::0, b : ?M::1, C : Type ⊢ ?M::0[lift:0:3] ≺ C
   Assumed: b
   Assumed: H
 Failed to solve
- ⊢ if ?M::0 (if (if ?M::3 (if a ⊥ ⊤) ⊤) ⊥ ⊤) ⊤ ≺ b
+ ⊢ ?M::0 ⇒ ?M::3 ∧ a ≺ b
-    Normalize
-     ⊢ if ?M::0 (?M::3 ∧ a) ⊤ ≺ b
     Substitution
-         ⊢ if ?M::0 ?M::1 ⊤ ≺ b
-            Normalize
      ⊢ ?M::0 ⇒ ?M::1 ≺ b
         (line: 20: pos: 18) Type of definition 't1' must be convertible to expected type.
         Assignment
@@ -342,9 +338,7 @@ Failed to solve
                  ⊢ ?M::1 ≈ Bool
                     Assumption 6
 Failed to solve
-a : Bool, b : Bool, H : ?M::2, H_a : ?M::6 ⊢ if (if a b ⊤) a ⊤ ≺ a
+a : Bool, b : Bool, H : ?M::2, H_a : ?M::6 ⊢ (a ⇒ b) ⇒ a ≺ a
-    Normalize
-    a : Bool, b : Bool, H : ?M::2, H_a : ?M::6 ⊢ (a ⇒ b) ⇒ a ≺ a
     Substitution
     a : Bool, b : Bool, H : ?M::2, H_a : ?M::6 ⊢ (a ⇒ b) ⇒ a ≺ ?M::5[lift:0:1]
         Substitution
@@ -371,8 +365,7 @@ a : Bool, b : Bool, H : ?M::2, H_a : ?M::6 ⊢ if (if a b ⊤) a ⊤ ≺ a
                         with arguments:
                             ?M::0
                             ?M::1
-                                λ H : ?M::2,
+                            λ H : ?M::2, DisjCases (EM a) (λ H_a : ?M::6, H) (λ H_na : ?M::7, NotImp1 (MT H H_na))
-                                  DisjCases (EM a) (λ H_a : ?M::6, H) (λ H_na : ?M::7, NotImp1 (MT H H_na))
                 Assignment
                 a : Bool, b : Bool ⊢ ?M::0 ≈ (a ⇒ b) ⇒ a
                     Destruct/Decompose

tests/lean/elab6.lean

@@ -1,2 +0,0 @@
-Theorem ForallIntro2 (A : (Type U)) (P : A -> Bool) (H : Pi x, P x) : forall x, P x :=
-        Abst (fun x, EqTIntro (H x))

tests/lean/exists4.lean.expected.out

@@ -17,7 +17,7 @@ Theorem T1 : ∃ x y z : N, P x y z :=
         a
         (ExistsIntro::explicit
              N
-             (λ x : N, if ((λ x::1 : N, if (P a x x::1) ⊥ ⊤) == (λ x : N, ⊤)) ⊥ ⊤)
+             (λ x : N, ¬ ∀ x::1 : N, ¬ P a x x::1)
              b
              (ExistsIntro::explicit N (λ z : N, P a b z) c H3))
 Theorem T2 : ∃ x y z : N, P x y z := ExistsIntro a (ExistsIntro b (ExistsIntro c H3))

tests/lean/interactive/elab6.lean

@@ -0,0 +1,9 @@
+(**
+-- The elaborator does not expand definitions marked as 'hidden'.
+-- To be able to prove the following theorem, we have to unmark the
+-- 'forall'
+local env = get_environment()
+set_hidden_flag(env, "forall", false)
+**)
+Theorem ForallIntro2 (A : (Type U)) (P : A -> Bool) (H : Pi x, P x) : forall x, P x :=
+        Abst (fun x, EqTIntro (H x))

tests/lean/interactive/elab6.lean.expected.out

@@ -1,3 +1,3 @@
-  Set: pp::colors
+#   Set: pp::colors
   Set: pp::unicode
   Proved: ForallIntro2

tests/lean/tst6.lean.expected.out

@@ -84,8 +84,8 @@ Theorem Example2 (a b c d : N)
                     a
                     b
                     c
-                    (Conjunct1::explicit (a == b) (b == c) H1)
+                    (Conjunct1::explicit (a == b) (eq::explicit N b c) H1)
-                    (Conjunct2::explicit (a == b) (b == c) H1))
+                    (Conjunct2::explicit (eq::explicit N a b) (b == c) H1))
                (Refl::explicit N b))
         (λ H1 : eq::explicit N a d ∧ eq::explicit N d c,
            CongrH::explicit
@@ -98,8 +98,8 @@ Theorem Example2 (a b c d : N)
                     a
                     d
                     c
-                    (Conjunct1::explicit (a == d) (d == c) H1)
+                    (Conjunct1::explicit (a == d) (eq::explicit N d c) H1)
-                    (Conjunct2::explicit (a == d) (d == c) H1))
+                    (Conjunct2::explicit (eq::explicit N a d) (d == c) H1))
                (Refl::explicit N b))
   Proved: Example3
   Set: lean::pp::implicit