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Proof step 13 from proof size-three (previous step)

a0 ∈ s, b ∈ s, 2 < # s, a ∈ s ⊦ # s = 0, b = a, a0 = b, a0 = a, ∃ a, b, c. a ∈ s ∧ b ∈ s ∧ c ∈ s ∧ a ≠ b ∧ a ≠ c ∧ b ≠ c

User applied rule exists right

And closed the goal!

Rule argument

pos: right 5 terms: b, a0, a abort: F discard: T computed: T

Additional information

Used terms: b, a0, a