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Proof step 9 from proof at-put-other-02 (previous step)

sks0 = [] ∨ (∀ st, r, sk, val, r0. (st[r -- sk, val][r0 -- sks0]) = st[r0 -- sks0] ∨ sk ∈ sks0) ⊦ (st[r -- sk, val][r0 -- (sk0 ' + sks0)]) = st[r0 -- (sk0 ' + sks0)], sk = sk0, sk ∈ sks0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 1:
⊦ sks ≠ [] → st[r -- (sk ' + sks)] = st[st[r -- sk] .refval -- sks]
Proof step 3:
¬ sks0 ≠ [], sks0 = [] ∨ (∀ st, r, sk, val, r0. (st[r -- sk, val][r0 -- sks0]) = st[r0 -- sks0] ∨ sk ∈ sks0) ⊦ (st[r -- sk, val][r0 -- (sk0 ' + sks0)]) = st[r0 -- (sk0 ' + sks0)], sk = sk0, sk ∈ sks0
Proof step 8:
sks0 ≠ [], sks0 = [] ∨ (∀ st, r, sk, val, r0. (st[r -- sk, val][r0 -- sks0]) = st[r0 -- sks0] ∨ sk ∈ sks0) ⊦ (st[r -- sk, val][(st[r -- sk, val][r0 -- sk0]) .refval -- sks0]) = st[r0 -- (sk0 ' + sks0)], sk = sk0, sk ∈ sks0

Rule argument

[crewrite-arg: lookup-rec-01 , F , [sks, sk, r, st] → [sks0, sk0, r0, st[r -- sk, val]] ]

Additional information

[[CONCRETESIMPRULES : ()], [EXTRATERMS : (sks0, sk0, r0, st[r -- sk, val])]]