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Proof step 9 from proof instore-rec-01 (previous step)

∀ r, sk, st. sks0 ≠ [] → (r -- (sk ' + sks0) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st) ⊦ sk0 + sks0 ≠ [] → (r -- (sk ' + (sk0 + sks0)) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- (sk0 + sks0) ∈ st)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
sks0 ≠ [] → (∀ r, sk, st. r -- (sk ' + sks0) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- sks0 ∈ st) ⊦ r -- (sk ' + sk0 ' + sks0) ∈ st ↔ r -- sk ∈ st ∧ st[r -- sk] .refval -- (sk0 ' + sks0) ∈ st

Rule argument

[emptyarg]

Additional information

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