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Proof step 3 from proof add-put-04 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ sks ≠ [] → st[r -- (sk ' + sks), val] = st[st[r -- sk] .refval -- sks, val]
Proof step 2:
⊦ sk ∈ fis → (addobj(r, class, fis, st)[r -- sk]) = fis[sk]

Rule argument

[speclemmaarg: , sks ≠ [] → st[r -- (sk ' + sks), val] = st[st[r -- sk] .refval -- sks, val] ,[val, sks, sk, r, st] → [val0, sks, sk, r, addobj(r, class1, fis, st)] ,F ,storefuns , ,put-rec-01 ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val0, sks, sk, r, addobj(r, class1, fis, st))]]