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Proof step 36 from proof addobj-addobj-switch (previous step)

⊦ r = r0, ∀ rk. (rk ∈ addobj(r, class, fis, addobj(r0, class0, fis0, st)) ↔ rk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[rk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[rk])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 35:
⊦ r = r0, (rk ∈ addobj(r, class, fis, addobj(r0, class0, fis0, st)) ↔ rk ∈ addobj(r0, class0, fis0, addobj(r, class, fis, st))) ∧ (addobj(r, class, fis, addobj(r0, class0, fis0, st))[rk]) = (addobj(r0, class0, fis0, addobj(r, class, fis, st))[rk])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]