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>
This commit fixes a problem exposed by t13.lean.
It has a theorem of the form:
Theorem T1 (A B : Bool) : A /\ B -> B /\ A :=
fun assumption : A /\ B,
let lemma1 := (show A by auto),
lemma2 := (show B by auto)
in (show B /\ A by auto)
When to_goal creates a goal for the metavariable associated with (show B /\ A by auto) it receives a context and proposition of the form
[ A : Bool, B : Bool, assumption : A /\ B, lemma1 := Conjunct1 assumption, lemma2 := Conjunct2 assumption ] |- B /\ A
The context_entries "lemma1 := Conjunct1 assumption" and "lemma2 := Conjunct2 assumption" do not have a domain (aka type).
Before this commit, to_goal would simply replace and references to "lemma1" and "lemma2" in "B /\ A" with their definitions.
Note that, "B /\ A" does not contain references to "lemma1" and "lemma2". Then, the following goal is created
A : Bool, B : Bool, assumption : A /\ B |- B /\ A
That is, the lemmas are not available when solving B /\ A.
Thus, the tactic auto produced the following (weird) proof for T1, where the lemmas are computed but not used.
Theorem T1 (A B : Bool) (assumption : A ∧ B) : B ∧ A :=
let lemma1 := Conjunct1 assumption,
lemma2 := Conjunct2 assumption
in Conj (Conjunct2 assumption) (Conjunct1 assumption)
This commit fixed that. It computes the types of "Conjunct1 assumption" and "Conjunct2 assumption", and creates the goal
A : Bool, B : Bool, assumption : A /\ B, lemma1 : A, lemma2 : B |- B /\ A
After this commit, the proof for theorem T1 is
Theorem T1 (A B : Bool) (assumption : A ∧ B) : B ∧ A :=
let lemma1 := Conjunct1 assumption,
lemma2 := Conjunct2 assumption
in Conj lemma2 lemma1
as expected.
Finally, this example suggests that the encoding
Theorem T1 (A B : Bool) : A /\ B -> B /\ A :=
fun assumption : A /\ B,
let lemma1 : A := (by auto),
lemma2 : B := (by auto)
in (show B /\ A by auto)
is more efficient than
Theorem T1 (A B : Bool) : A /\ B -> B /\ A :=
fun assumption : A /\ B,
let lemma1 := (show A by auto),
lemma2 := (show B by auto)
in (show B /\ A by auto)
Signed-off-by: Leonardo de Moura <leonardo@microsoft.com>
