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Proof step 3 from proof sw-01 (previous step)

a0 = b ⊦ (a0 ∈ st -- a -- b ↔ a0 ∈ st -- b -- a) ∧ (st -- a -- b)[a0] = (st -- b -- a)[a0]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ (b ∈ st -- a -- b ↔ b ∈ st -- b -- a) ∧ (∅)[b] = (st -- b -- a)[b]

Rule argument

Additional information

Used simplifier rules: lookupisempty: ⊦ (st -- a)[a] = (∅)[a];