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Proof step 13 from proof in-list-subset (previous step)

sks0 ≠ [], st ⊆ st0, r -- (sk ' + sks0) ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → r0 -- sks0 ∈ st2 ⊦ sks0 = [], r -- sk ∈ st0 ∧ st0[r -- sk] .refval -- sks0 ∈ st0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ sks ≠ [] → (r -- (sk ' + sks) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sks ∈ st)
Proof step 12:
sks0 ≠ [], sks0 ≠ [], st ⊆ st0, r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → r0 -- sks0 ∈ st2 ⊦ sks0 = [], r -- sk ∈ st0 ∧ st0[r -- sk] .refval -- sks0 ∈ st0

Rule argument

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

Additional information

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