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Proof step 8 from proof del-ne (previous step)

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

Heuristic quantifier closing applied rule all left

And obtained the following new goals

Proof step 9:
a1 ∈ s, b ∈ s, a ∈ s, ∀ a0. a0 = a ∨ ¬ a0 ∈ s, b = a ∨ ¬ b ∈ s ⊦ # s = 0, # s = 1, a1 = b

Rule argument

pos: left 4 terms: b

Additional information

Used terms: b