lean4-htt/tests/lean/run/adamTC2.lean
2022-03-12 18:44:52 -08:00

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inductive Expr where
| nat : Nat → Expr
| plus : Expr → Expr → Expr
| bool : Bool → Expr
| and : Expr → Expr → Expr
inductive Ty where
| nat
| bool
deriving DecidableEq
inductive HasType : Expr → Ty → Prop
| nat : HasType (.nat v) .nat
| plus : HasType a .nat → HasType b .nat → HasType (.plus a b) .nat
| bool : HasType (.bool v) .bool
| and : HasType a .bool → HasType b .bool → HasType (.and a b) .bool
inductive Maybe (p : α → Prop) where
| unknown
| found : (a : α) → p a → Maybe p
notation "{{ " x " | " p " }}" => Maybe (fun x => p)
def Expr.typeCheck (e : Expr) : {{ ty | HasType e ty }} :=
match e with
| nat .. => .found .nat .nat
| bool .. => .found .bool .bool
| plus a b =>
match a.typeCheck, b.typeCheck with
| .found .nat h₁, .found .nat h₂ => .found .nat (.plus h₁ h₂)
| _, _ => .unknown
| and a b =>
match a.typeCheck, b.typeCheck with
| .found .bool h₁, .found .bool h₂ => .found .bool (.and h₁ h₂)
| _, _ => .unknown
theorem HasType.det (h₁ : HasType e t₁) (h₂ : HasType e t₂) : t₁ = t₂ := by
cases h₁ <;> cases h₂ <;> rfl
-- TODO: for simplifying the following proof we need: ematching for forward reasoning, and `match` blast for case analysis
theorem Expr.typeCheck_complete {e : Expr} : e.typeCheck = .unknown → ¬ HasType e t := by
induction e with simp [typeCheck]
| plus a b iha ihb =>
revert iha ihb
cases typeCheck a <;> cases typeCheck b <;> simp <;> intros <;> intro h <;> cases h <;> try contradiction
rename_i ty₁ _ ty₂ _ h _ _
cases ty₁ <;> cases ty₂ <;> simp at h
. have := HasType.det HasType b Ty.bool HasType b Ty.nat; contradiction
. have := HasType.det HasType a Ty.bool HasType a Ty.nat; contradiction
. have := HasType.det HasType a Ty.bool HasType a Ty.nat; contradiction
| and a b iha ihb =>
revert iha ihb
cases typeCheck a <;> cases typeCheck b <;> simp <;> intros <;> intro h <;> cases h <;> try contradiction
rename_i ty₁ _ ty₂ _ h _ _
cases ty₁ <;> cases ty₂ <;> simp at h
. have := HasType.det HasType b Ty.bool HasType b Ty.nat; contradiction
. have := HasType.det HasType a Ty.bool HasType a Ty.nat; contradiction
. have := HasType.det HasType b Ty.bool HasType b Ty.nat; contradiction
instance (e : Expr) (t : Ty) : Decidable (HasType e t) :=
match h' : e.typeCheck with
| .found t' ht' =>
if heq : t = t' then
isTrue (heq ▸ ht')
else
isFalse fun ht => heq (HasType.det ht ht')
| .unknown => isFalse (Expr.typeCheck_complete h')