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Proof step 16 from proof okclass-03 (conclusion)

⊦ sk ≠ _type → (okclass(val, class1, st[r0 -- sk, val0]) ↔ okclass(val, class1, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ okclass(val, class1, st) ↔ reftypep(val, st) ∧ (val .refval ≠ jvmref → val .refval -- _type ∈ st ∧ st[val .refval -- _type] = typeval(mkclasstype(class1)))
Proof step 15:
⊦ sk = _type, reftypep(val, st[r0 -- sk, val0]) ∧ (val .refval ≠ jvmref → val .refval -- _type ∈ st[r0 -- sk, val0] ∧ st[val .refval -- _type] = typeval(mkclasstype(class1))) ↔ okclass(val, class1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sk ≠ sk0 → (st[r -- sk, val][r0 -- sk0]) = st[r0 -- sk0]])]]