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Proof step 1 from proof sw (conclusion)

⊦ a ≠ b → st[b, d] -- a = (st -- a)[b, d]

Heuristic simplifier applied rule simplifier

And obtained the following new goals

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

Rule argument

Additional information

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