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Proof step 9 from proof size-two-03 (
previous step
)
# s ≠ 0 → a0 ≠ b ∧ (a0 = a2 ∨ a0 ∈ s) ∧ (b = a2 ∨ b ∈ s) ⊦ a = a2, a2 ∈ s, a ∈ s, ∃ a1, b0. a1 ≠ b0 ∧ (a1 = a ∨ a1 = a2 ∨ a1 ∈ s) ∧ (b0 = a ∨ b0 = a2 ∨ b0 ∈ s)
User applied rule exists right
And closed the goal!
Rule argument
pos: right 4 terms: a, a2 abort: F discard: F computed: T
Additional information
Used terms: a, a2