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Proof step 7 from proof del-ne (previous step)

a0 ∈ s, b ∈ s, a1 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, a0 = a ∨ ¬ a0 ∈ s ⊦ # s = 1, # s = 0, a1 = b

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
a ∈ s, b ∈ s, a1 ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s ⊦ # s = 1, # s = 0, a1 = b

Rule argument

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