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Proof step 10 from proof at-put-same (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks]
Proof step 9:
sks0 ≠ [], ∀ st0, r0, val0. (st0[r0 -- sks0, val0][r0 -- sks0]) = val0 ⊦ sks0 = [], (st[st[r -- sk] .refval -- sks0, val][(st[st[r -- sk] .refval -- sks0, val][r -- sk]) .refval -- sks0]) = val, sk ∈ sks0, dups(sks0)

Rule argument

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

Additional information

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