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Proof step 8 from proof put-put-switch (previous step)

sks ≠ [], ∀ st, r, sk, val, r0, val0. (st[r -- sk, val][r0 -- sks, val0]) = (st[r0 -- sks, val0][r -- sk, val]) ∨ sk ∈ sks ⊦ sk = sk0, sks = [], (st[r -- sk, val][st[r0 -- sk0] .refval -- sks, val0]) = (st[st[r0 -- sk0] .refval -- sks, val0][r -- sk, val]), sk ∈ sks

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
∀ st, r, sk, val, r0, val0. (st[r -- sk, val][r0 -- sks, val0]) = (st[r0 -- sks, val0][r -- sk, val]) ∨ sk ∈ sks ⊦ sks = [], (st[r -- sk, val][st[r0 -- sk0] .refval -- sks, val0]) = (st[st[r0 -- sk0] .refval -- sks, val0][r -- sk, val]), sk = sk0, sk ∈ sks

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ()]]