Helper type for implementing discharge?'
- proved: Lean.Meta.Simp.DischargeResult
- notProved: Lean.Meta.Simp.DischargeResult
- maxDepth: Lean.Meta.Simp.DischargeResult
- failedAssign: Lean.Meta.Simp.DischargeResult
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- Lean.Meta.Simp.instDecidableEqDischargeResult x y = if h : x.toCtorIdx = y.toCtorIdx then isTrue ⋯ else isFalse ⋯
Wrapper for invoking discharge?
method. It checks for maximum discharge depth, create trace nodes, and ensure
the generated proof was successfully assigned to x
.
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Remark: the parameter tag is used for creating trace messages. It is irrelevant otherwise.
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For (← getConfig).index := true
, use discrimination tree structure when collecting simp
theorem candidates.
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For (← getConfig).index := false
, Lean 3 style simp
theorem retrieval.
Only the root symbol is taken into account. Most of the structure of the discrimination tree is ignored.
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- Lean.Meta.Simp.rewrite?.inErasedSet erased thm = Lean.PersistentHashSet.contains erased thm.origin
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Given a match-application e
with MatcherInfo
info
, return some result
if at least of one of the discriminants has been simplified.
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Discharge procedure for the ground/symbolic evaluator.
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Try to unfold ground term in the ground/symbolic evaluator.
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- Lean.Meta.Simp.postSEval s = Lean.Meta.Simp.rewritePost >> Lean.Meta.Simp.userPostSimprocs s >> Lean.Meta.Simp.sevalGround
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Invoke ground/symbolic evaluator from simp
.
It uses the seval
theorems and simprocs.
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Try to unfold ground term in the ground/symbolic evaluator.
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Return true if e
is of the form (x : α) → ... → s = t → ... → False
Recall that this kind of proposition is generated by Lean when creating equations for
functions and match-expressions with overlapping cases.
Example: the following match
-expression has overlapping cases.
def f (x y : Nat) :=
match x, y with
| Nat.succ n, Nat.succ m => ...
| _, _ => 0
The second equation is of the form
(x y : Nat) → ((n m : Nat) → x = Nat.succ n → y = Nat.succ m → False) → f x y = 0
The hypothesis (n m : Nat) → x = Nat.succ n → y = Nat.succ m → False
is essentially
saying the first case is not applicable.
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Tries to solve e
using unifyEq?
.
It assumes that isEqnThmHypothesis e
is true
.
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Discharges assumptions of the form ∀ …, a = b
using rfl
. This is particularly useful for higher
order assumptions of the form ∀ …, e = ?g x y
to instaniate a parameter g
even if that does not
appear on the lhs of the rule.
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