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Proof step 14 from proof put-put-switch (conclusion)

⊦ ¬ sk ∈ sks0 ∧ sks0 ≠ [] → (st[r -- sk, val][r0 -- sks0, val0]) = (st[r0 -- sks0, val0][r -- sk, val])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
⊦ sks0 = [], (st[r -- sk, val][r0 -- sks0, val0]) = (st[r0 -- sks0, val0][r -- sk, val]), sk ∈ sks0

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]