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Proof step 3 from proof add-put-03 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 1:
⊦ addobj(r, class1, (sk × val) ', st) = (addobj(r, class1, [], st)[r -- sk, val])
Proof step 2:
⊦ addobj(r, class1, [], st) = st[r -- _type, typeval(mkclasstype(class1))]

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks ≠ [] → (st[r -- sk, val] [r -- (sk ' + sks), val0 ] ) = (st[r -- sk, val] [val .refval -- sks, val0 ] )])]]