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Proof step 7 from proof lookup-last-rec (previous step)

sks0 + sk ' ≠ [], sks0 = [] ∨ (∀ st, r, sk. st[r -- (sks0 + sk ')] = st[st[r -- sks0] .refval -- sk]) ⊦ st[st[r -- sk0] .refval -- (sks0 + sk ')] = st[st[r -- (sk0 ' + sks0)] .refval -- sk]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st[r -- sk '] = st[r -- sk]
Proof step 6:
∀ st, r, sk. st[r -- (sks0 + sk ')] = st[st[r -- sks0] .refval -- sk] ⊦ st[st[r -- sk0] .refval -- (sks0 + sk ')] = st[st[r -- (sk0 ' + sks0)] .refval -- sk], sks0 = []

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sks + [] = sks], [CSIMPSEQ : ⊦ [] + sks = sks], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks + sk ' ≠ []])]]