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Proof step 5 from proof sub-del-any (previous step)

⊦ (∀ a. #oc(a, x) ≤ #oc(a, y)) ↔ (∀ a0. #oc(a0, x -l a) ≤ #oc(a0, y -l a)) ∧ #oc(a, x) ≤ #oc(a, y)

User applied rule case distinction

And obtained the following new goals

Proof step 6:
∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ (∀ a0. #oc(a0, x -l a) ≤ #oc(a0, y -l a)) ∧ #oc(a, x) ≤ #oc(a, y)
Proof step 15:
(∀ a0. #oc(a0, x -l a) ≤ #oc(a0, y -l a)) ∧ #oc(a, x) ≤ #oc(a, y) ⊦ ∀ a. #oc(a, x) ≤ #oc(a, y)

Rule argument

pos: right 1

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