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Proof step 6 from proof isobj-union (conclusion)

⊦ ¬ r ∈ st .refs → (is_obj(r, class, sks, st ∪ st0) ↔ is_obj(r, class, sks, st0))

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ is_obj(r, class, sks, st ∪ st0) ↔ is_obj(r, class, sks, st0), r ∈ st .refs

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]