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Proof step 4 from proof addobj-at-other (conclusion)

⊦ r ≠ r0 ∧ sks ≠ [] → (addobj(r, class, fis, st)[r0 -- (sk ' + sks)]) = (addobj(r, class, fis, st)[st[r0 -- sk] .refval -- sks])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ r = r0, sks = [], (addobj(r, class, fis, st)[r0 -- (sk ' + sks)]) = (addobj(r, class, fis, st)[st[r0 -- sk] .refval -- sks])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]