98bd162ad4
This PR adds support for closing goals using `match`-expression
conditions that are known to be true in the `grind` tactic state.
`grind` can now solve goals such as:
```lean
def f : List Nat → List Nat → Nat
| _, 1 :: _ :: _ => 1
| _, _ :: _ => 2
| _, _ => 0
example : z = a :: as → y = z → f x y > 0
```
Without `grind`, we would use the `split` tactic. The first two goals,
corresponding to the first two alternatives, are closed using `simp`,
and the the third using the `match`-expression condition produced by
`split`. The proof would proceed as follows.
```lean
example : z = a :: as → y = z → f x y > 0 := by
intros
unfold f
split
next => simp
next => simp
next h =>
/-
...
_ : z = a :: as
_ : y = z
...
h : ∀ (head : Nat) (tail : List Nat), y = head :: tail → False
|- 0 > 0
-/
subst_vars
/-
...
h : ∀ (head : Nat) (tail : List Nat), a :: as = head :: tail → False
|- 0 > 0
-/
have : False := h a as rfl
contradiction
```
Here is the same proof using `grind`.
```lean
example : z = a :: as → y = z → f x y > 0 := by
grind [f.eq_def]
```
7 lines
168 B
Text
7 lines
168 B
Text
def f : List Nat → List Nat → Nat
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| _, 1 :: _ :: _ => 1
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| _, _ :: _ => 2
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| _, _ => 0
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example : z = a :: as → y = z → f x y > 0 := by
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grind [f.eq_def]
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