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Proof step 8 from proof lookup-append (previous step)

sks1 + sks0 ≠ [], ∀ sks0, st, r. sks0 = [] ∨ st[r -- (sks1 + sks0)] = st[st[r -- sks1] .refval -- sks0] ⊦ sks1 = [], st[st[r -- sk] .refval -- (sks1 + sks0)] = st[st[st[r -- sk] .refval -- sks1] .refval -- sks0], sks0 = []

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ sks0, st, r. sks0 = [] ∨ st[r -- (sks1 + sks0)] = st[st[r -- sks1] .refval -- sks0] ⊦ sks1 + sks0 = [], sks0 = [], st[st[r -- sk] .refval -- (sks1 + sks0)] = st[st[st[r -- sk] .refval -- sks1] .refval -- sks0], sks1 = []

Rule argument

[emptyarg]

Additional information

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