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Proof step 4 from proof perm-all (previous step)

⊦ (∀ a. #oc(a, x) ≤ #oc(a, y)) ∧ (∀ a. #oc(a, y) ≤ #oc(a, x)) ↔ (∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y))

User applied rule case distinction

And obtained the following new goals

Proof step 5:
(∀ a. #oc(a, x) ≤ #oc(a, y)) ∧ (∀ a. #oc(a, y) ≤ #oc(a, x)) ⊦ ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y)
Proof step 9:
∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y) ⊦ (∀ a. #oc(a, x) ≤ #oc(a, y)) ∧ (∀ a. #oc(a, y) ≤ #oc(a, x))

Rule argument

pos: right 1

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