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Proof step 12 from proof at-put-same (previous step)

∀ st0, r0, val0. (st0[r0 -- sks0, val0][r0 -- sks0]) = val0 ⊦ (st[r -- (sk ' + sks0), val][r -- (sk ' + sks0)]) = val, sks0 = [], dups(sks0), sk ∈ sks0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ sks ≠ [] → st[r -- (sk ' + sks), val] = st[st[r -- sk] .refval -- sks, val]
Proof step 11:
sks0 ≠ [], ∀ st0, r0, val0. (st0[r0 -- sks0, val0][r0 -- sks0]) = val0 ⊦ (st[st[r -- sk] .refval -- sks0, val][r -- (sk ' + sks0)]) = val, sks0 = [], dups(sks0), sk ∈ sks0

Rule argument

[crewrite-arg: put-rec-01 , F , [val, sks, sk, r, st] → [val, sks0, sk, r, st] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (val, sks0, sk, r, st)]]