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Proof step 5 from proof at-put-first-04 (conclusion)

⊦ sks0 ≠ [] → (st[r -- (sk ' + sk0 '), val][st[r -- sk] .refval -- (sk0 ' + sks0)]) = (st[r -- (sk ' + sk0 '), val][val .refval -- sks0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ sks0 = [], (st[r -- (sk ' + sk0 '), val][st[r -- sk] .refval -- (sk0 ' + sks0)]) = (st[r -- (sk ' + sk0 '), val][val .refval -- sks0])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]