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Proof step 5 from proof addobj-at-this-01 (conclusion)

⊦ (addobj(r, class, (sk × val) ', st)[r -- (sk ' + sk0 ' + sks)]) = (addobj(r, class, (sk × val) ', st)[val .refval -- (sk0 ' + sks)])

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))]
Proof step 4:
⊦ (st[r -- _type, typeval(mkclasstype(class))][r -- sk, val][r -- (sk ' + sk0 ' + sks)]) = (st[r -- _type, typeval(mkclasstype(class))][r -- sk, val][val .refval -- (sk0 ' + sks)])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]