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Proof step 20 from proof size-three-02 (previous step)

⊦ a = a0, a0 = a1, a = a1, ∃ a2, b, c. a2 ≠ b ∧ a2 ≠ c ∧ b ≠ c ∧ ∅ ++ a1 ++ a0 ++ a = ∅ ++ a2 ++ b ++ c

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 4 terms: a1, a0, a

Additional information

Used terms: a1, a0, a