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Proof step 4 from proof add-put-04 (conclusion)

⊦ sk ∈ fis ∧ sks ≠ [] → (addobj(r, class1, fis, st)[r -- (sk ' + sks), val0]) = (addobj(r, class1, fis, st)[fis[sk] .refval -- sks, val0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
sk ∈ fis ⊦ sks = [], (addobj(r, class1, fis, st)[r -- (sk ' + sks), val0]) = (addobj(r, class1, fis, st)[fis[sk] .refval -- sks, val0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]