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Proof step 7 from proof in-put (previous step)

sks0 = sks0 .first + [] ⊦ sks0 = [], r0 -- sk ∈ st[r -- (sk0 ' + sks0), val] ↔ r0 -- sk ∈ st ∨ r0 = st[r -- (sk0 ' + sks0 .butlast)] .refval ∧ sk = sks0 .last

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
sks0 = sks0 .first ' ⊦ sks0 = [], r0 -- sk ∈ st[r -- (sk0 ' + sks0), val] ↔ r0 -- sk ∈ st ∨ r0 = st[r -- (sk0 ' + sks0 .butlast)] .refval ∧ sk = sks0 .first

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ sks = sks .first ' → sks .last = sks .first], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks + [] = sks], [CSIMPSEQ : ⊦ sk + sks = sk ' + sks])]]