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Proof step 1 from proof sw-01 (conclusion)

⊦ st -- a -- b = st -- b -- a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
⊦ (a0 ∈ st -- a -- b ↔ a0 ∈ st -- b -- a) ∧ (st -- a -- b)[a0] = (st -- b -- a)[a0]

Rule argument

Additional information

Used simplifier rules: Extension: ⊦ st1 = st2 ↔ (∀ a. (a ∈ st1 ↔ a ∈ st2) ∧ st1[a] = st2[a]);