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Proof step 6 from proof in-put (previous step)

sks0 = sks0 .first ' ⊦ sks0 = [], r0 -- sk ∈ st[r -- (sk0 ' + sks0), val] ↔ r0 -- sk ∈ st ∨ r0 = st[r -- (sk0 ' + sks0 .butlast)] .refval ∧ sk = sks0 .first

Heuristic elimination applied rule insert elim lemma

And obtained the following new goals

Proof step 5:
sks0 = sk1 + sks, sks0 = sk1 ' ⊦ sks0 = [], r0 -- sk ∈ st[r -- (sk0 ' + sks0), val] ↔ r0 -- sk ∈ st ∨ r0 = st[r -- (sk0 ' + sks0 .butlast)] .refval ∧ sk = sk1

Rule argument

[speclemmaarg: sks ≠ [] ,sk = sks .first ∧ sks0 = sks .rest ↔ sks = sk + sks0 ,[sks, sk, sks0] → [sks0, sk1, sks] ,T ,list-data ,storekeys ,elim-list-c ]

Additional information

[[CONCRETESIMPRULES : ([CSIMPELIM : sks ≠ [] ⊦ sk = sks .first ∧ sks0 = sks .rest ↔ sks = sk + sks0])], [EXTRATERMS : (sks0, sk1, sks)]]