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Proof step 11 from proof Diff-del-rw (previous step)

⊦ (a0 ∈ st -- a ∧ ¬ a0 ∈ s ↔ a0 ≠ a ∧ a0 ∈ st ∧ ¬ a0 ∈ s) ∧ (a0 ∈ st -- a ∧ ¬ a0 ∈ s → (st -- a)[a0] = ((st \ s) -- a)[a0])

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
⊦ (¬ a0 ∈ s → a0 ∈ st -- a → a0 ≠ a ∧ a0 ∈ st) ∧ (a0 ∈ st -- a ∧ ¬ a0 ∈ s → (st -- a)[a0] = ((st \ s) -- a)[a0])

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ st -- b ↔ a ≠ b ∧ a ∈ st; (ws)