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Proof step 13 from proof in-put (previous step)

m1 < # x0, a ∈ x0, ∀ a1, m0, b0. a1 ∈ x0 ∧ #oc(a1, x0) ≠ 1 ∧ m0 < # x0 → a1 ∈ x0[m0, b0] ⊦ #oc(a, x0) = 1, a = a0, x0 = [], a ∈ x0[m1, b]

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a, m1, b

Additional information

Used terms: a, m1, b