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Proof step 5 from proof diff (previous step)

∀ a. #oc(a, y0) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, y0), ∀ a. #oc(a, x0) ≤ #oc(a, x), ∀ a. #oc(a, x) ≤ #oc(a, x0) ⊦ #oc(a, x \ y) ≤ #oc(a, x0 \ y0) ∧ #oc(a0, x0 \ y0) ≤ #oc(a0, x \ y)

User applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

Additional information