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Proof step 2 from proof app-01 (previous step)

∀ a. #oc(a, x) ≤ #oc(a, z), ∀ a. #oc(a, z) ≤ #oc(a, x) ⊦ #oc(a, x) ≤ #oc(a, z) ∧ #oc(a0, z) ≤ #oc(a0, x)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
∀ a. #oc(a, x) ≤ #oc(a, z), ∀ a. #oc(a, z) ≤ #oc(a, x) ⊦ #oc(a, x) ≤ #oc(a, z)
Proof step 5:
∀ a. #oc(a, x) ≤ #oc(a, z), ∀ a. #oc(a, z) ≤ #oc(a, x) ⊦ #oc(a0, z) ≤ #oc(a0, x)

Rule argument

pos: right 1

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