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Proof step 3 from proof sw (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
⊦ a0 = a, a = b, (a0 ∈ st[b, d] ↔ a0 ∈ (st -- a)[b, d]) ∧ st[b, d][a0] = (st -- a)[b, d][a0]

Rule argument

Additional information

Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;%Del-at-strong: a ≠ b ⊦ (st -- b)[a] = st[a];