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Proof step 5 from proof put-put-switch (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ st[r -- sk ', val] = st[r -- sk, val]
Proof step 4:
⊦ sk = sk0, (st[r -- sk, val][r0 -- sk0, val0]) = (st[r0 -- sk0, val0][r -- sk, val])

Rule argument

[emptyarg]

Additional information

[[CONCRETESIMPRULES : ([CSIMPSEQ : ⊦ ¬ sk ∈ []], [CSIMPSEQ : ⊦ (sks + sks0) + sks1 = sks + sks0 + sks1], [CSIMPSEQ : ⊦ sks + [] = sks])]]