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Proof step 4 from proof notin-del-notin (previous step)

⊦ ∀ a0. (a0 ∈ st -- a ↔ a0 ∈ st) ∧ (st -- a)[a0] = st[a0], a ∈ st

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
⊦ a ∈ st, (a0 ∈ st -- a → a0 ∈ st) ∧ (st -- a)[a0] = st[a0]

Rule argument

Additional information

Used simplifier rules: Del-in: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st;