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Proof step 3 from proof addobj-put-switch (conclusion)

⊦ r ≠ r0 → addobj(r, class1, fis, st[r0 -- sk, val]) = (addobj(r, class1, fis, st)[r0 -- sk, val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ r = r0, addobj(r, class1, fis, st[r0 -- sk, val]) = (addobj(r, class1, fis, st)[r0 -- sk, val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]