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Proof step 5 from proof delapp (previous step)

(∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a)) ∧ (∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y)) ⊦ ¬ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
a ∈ y, ∀ a0. #oc(a0, y) ≤ #oc(a0, y -1l a), ∀ a0. #oc(a0, y -1l a) ≤ #oc(a0, y) ⊦

Rule argument

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