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Proof step 6 from proof classof-01 (conclusion)

sk ≠ _type ⊦ classof(r, class1, st[r0 -- sk, val]) ↔ classof(r, class1, st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ sk = _type, classof(r, class1, st[r0 -- sk, val]) ↔ classof(r, class1, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]