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Proof step 4 from proof eq-02 (previous step)

∀ a0. a0 = a ∨ a0 ∈ s ↔ a0 = b ⊦ ¬ a0 ∈ s, a ∈ s

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
a0 ∈ s, ∀ a0. a0 = a ∨ a0 ∈ s ↔ a0 = b ⊦ a ∈ s

Rule argument

Additional information

Used simplifier rules: ne-01: ⊦ a ∈ s → s ≠ ∅;%ne-02: ⊦ a ∈ s → (s = ∅ ↔ false);