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Proof step 5 from proof size-two-04 (previous step)

a0 ∈ s, a ∈ s ⊦ # s = 0, # s = 1, a0 = a, ∃ b. b ∈ s ∧ a ≠ b

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: a0

Additional information

Used terms: a0