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Proof step 14 from proof lookup-list-intro-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
sks1 = [] ∨ (∀ sks0, st, r. sks0 = [] ∨ st[st[r -- sks1] .refval -- sks0] = st[r -- (sks1 + sks0)]) ⊦ sks0 = [], st[st[r -- (sk ' + sks1)] .refval -- sks0] = st[r -- ((sk ' + sks1) + sks0)]

Rule argument

[emptyarg]

Additional information

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