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Proof step 5 from proof size-two-03 (previous step)

b ∈ s0, a0 ∈ s0, a ∈ s0 ⊦ a0 = b, # s0 = 1, # s0 = 0, ∃ a1, b0. a1 ≠ b0 ∧ a1 ∈ s0 ∧ b0 ∈ s0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: a0, b

Additional information

Used terms: a0, b