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Proof step 12 from proof violations-exists (previous step)

less#(a, x0) = 0, violations(x0) = 0, m + 1 < # x0, x0[m + 1] < x0[m] ⊦ x0 = [], ∃ n0. x0[n0 + 1] < x0[n0] ∧ n0 + 1 < # x0

Heuristic quantifier closing applied rule exists right

And closed the goal!

Rule argument

pos: right 2 terms: m

Additional information

Used terms: m