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Proof step 2 from proof in (previous step)

a ∈ y, ∀ a. #oc(a, x) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, x) ⊦ a ∈ x

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%lez-01: ⊦ m ≤ 0 ↔ m = 0; from specification nat Used terms: a