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Proof step 3 from proof del-two (previous step)

a1 ∈ s, a0 ∈ s, a4 ∈ s, ∀ a4. a4 = b ∨ a4 = a ∨ ¬ a4 ∈ s ⊦ a3 ∈ s ↔ a3 ∈ ∅ ++ a ++ b, a2 ∈ s ↔ a2 = b, a1 ∈ s ↔ a1 = a

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information