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Proof step 10 from proof del-e (previous step)

⊦ (∀ a0. a0 = b ∨ ¬ a0 ∈ s ++ a) ↔ (∀ a0. ¬ a0 ∈ s) ∧ a = b, a = b, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
∀ a0. a0 = b ∨ a0 ≠ a ∧ ¬ a0 ∈ s ⊦ a = b, a ∈ s

Rule argument

Additional information

Used simplifier rules: ⊦ a ∈ s ++ b ↔ a = b ∨ a ∈ s; (s)