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Proof step 4 from proof ad (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 5:
#oc(a0, (x + y) -1l a) < #oc(a0, x) + #oc(a0, y -1l a), a ∈ y ⊦ #oc(a0, x) + #oc(a0, y -1l a) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)