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Proof step 6 from proof eq-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
a0 ∈ s, ∀ a. ¬ a ∈ s ⊦ b ∈ s

Rule argument

Additional information