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Proof step 9 from proof lookup-list-intro-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
∀ st, r, sk0. st[st[r -- sks0] .refval -- sk0] = st[r -- (sks0 + sk0 ')] ⊦ sks0 = [], st[st[st[r -- sk] .refval -- sks0] .refval -- sk0] = st[st[r -- sk] .refval -- (sks0 + sk0 ')]

Rule argument

[emptyarg]

Additional information

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