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Proof step 4 from proof at-put-first-01 (conclusion)

⊦ ¬ sk ∈ sks ∧ sks ≠ [] → (st[r -- sk, val][r -- (sk ' + sks)]) = st[val .refval -- sks]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
⊦ sks = [], (st[r -- sk, val][r -- (sk ' + sks)]) = st[val .refval -- sks], sk ∈ sks

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]