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Proof step 23 from proof isobj-addobj (conclusion)

⊦ r ≠ r0 → (is_obj(r, class, sks, addobj(r0, class0, fis, st)) ↔ is_obj(r, class, sks, st))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
⊦ r = r0, is_obj(r, class, sks, addobj(r0, class0, fis, st)) ↔ is_obj(r, class, sks, st)

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]