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Proof step 3 from proof at-put-same-02 (conclusion)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ sks = [], (st[r -- (sks + sk '), val][st[r -- sks] .refval -- sk]) = val

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]