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Proof step 7 from proof haskeys-addobj (conclusion)

⊦ r ≠ r0 → (hasKeys(r, sks, addobj(r0, class, fis, st)) ↔ hasKeys(r, sks, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
⊦ r = r0, hasKeys(r, sks, addobj(r0, class, fis, st)) ↔ hasKeys(r, sks, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]