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Proof step 11 from proof sub-del-any (previous step)

a0 ≠ a, ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ #oc(a0, x) ≤ #oc(a0, y -l a) ∧ #oc(a, x) ≤ #oc(a, y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
∀ a. #oc(a, x) ≤ #oc(a, y) ⊦ a0 = a, #oc(a0, x) ≤ #oc(a0, y) ∧ #oc(a, x) ≤ #oc(a, y)

Rule argument

Additional information

Used simplifier rules: ocs-del-length: ⊦ a ≠ b → #oc(a, x -l b) = #oc(a, x);