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Proof step 10 from proof diff (previous step)

a ∈ y0, a ∈ x, ∀ a. #oc(a, x) ≤ #oc(a, x0), ∀ a. #oc(a, x0) ≤ #oc(a, x), ∀ a. #oc(a, y) ≤ #oc(a, y0), ∀ a. #oc(a, y0) ≤ #oc(a, y) ⊦ a ∈ y

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 6 terms: a

Additional information

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