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Proof step 2 from proof ad (previous step)

a ∈ y ⊦ #oc(a0, x) + #oc(a0, y -1l a) ≤ #oc(a0, (x + y) -1l a) ∧ #oc(a1, (x + y) -1l a) ≤ #oc(a1, x) + #oc(a1, y -1l a)

User applied rule case distinction

And obtained the following new goals

Proof step 3:
a ∈ y ⊦ #oc(a0, x) + #oc(a0, y -1l a) ≤ #oc(a0, (x + y) -1l a)
Proof step 6:
a ∈ y ⊦ #oc(a1, (x + y) -1l a) ≤ #oc(a1, x) + #oc(a1, y -1l a)

Rule argument

pos: right 1

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