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Proof step 15 from proof addobj-addobj-switch (previous step)

¬ r1 ≠ r ⊦ r = r0, (r1 -- sk ∈ addobj(r, class, fis, addobj(r0, class0, fis0, st)) ↔ r1 -- sk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[r1 -- sk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[r1 -- sk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]