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Proof step 10 from proof shift-02 (previous step)

∀ a. #oc(a, x0) + #oc(a, x) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, x0) + #oc(a, x) ⊦ #oc(a0, y) ≤ #oc(a0, x) + #oc(a0, x0)

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 2 terms: a0

Additional information

Used simplifier rules: c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: a0