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Proof step 6 from proof app-02 (previous step)

#oc(a0, y) < #oc(a0, z), a0 ∈ z, ∀ a. #oc(a, z) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, z) ⊦

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: a0

Additional information

Used simplifier rules: le-ls-simp: ⊦ m ≤ n → (n < m ↔ false); from specification nat Used terms: a0