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Proof step 7 from proof subrec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 8:
⊦ (∀ a0. #oc(a0, a ') + #oc(a0, x) ≤ #oc(a0, y)) ↔ a ∈ y ∧ (∀ a0. #oc(a0, x) ≤ #oc(a0, y -1l a))

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%a: ⊦ (x + y) + z = x + y + z; from specification list