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Proof step 8 from proof subrec (previous step)

⊦ (∀ a0. #oc(a0, a ') + #oc(a0, x) ≤ #oc(a0, y)) ↔ a ∈ y ∧ (∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a))

Heuristic pl case distinction applied rule case distinction

And obtained the following new goals

Proof step 9:
∀ a0. #oc(a0, a ') + #oc(a0, x) ≤ #oc(a0, y) ⊦ a ∈ y ∧ (∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a))
Proof step 15:
a ∈ y, ∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a) ⊦ ∀ a0. #oc(a0, a ') + #oc(a0, x) ≤ #oc(a0, y)

Rule argument

pos: right 1

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