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Proof step 14 from proof lookup-subset (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks]
Proof step 13:
sks0 ≠ [], st ⊆ st0, r -- sk ∈ st, st[r -- sk] .refval -- sks0 ∈ st, r ∈ st, ∀ r0, st1, st2. r0 -- sks0 ∈ st1 ∧ st1 ⊆ st2 → st2[r0 -- sks0] = st1[r0 -- sks0] ⊦ st0[st0[r -- sk] .refval -- sks0] = st[r -- (sk ' + sks0)], sks0 = []

Rule argument

[crewrite-arg: lookup-rec-01 , F , [sks, sk, r, st] → [sks0, sk, r, st0] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks0, sk, r, st0)]]