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Proof step 5 from proof c (previous step)

∀ a. #oc(a, x) ≤ #oc(a, y) + #oc(a, z), ∀ a. #oc(a, y) + #oc(a, z) ≤ #oc(a, x) ⊦ #oc(a0, y) + #oc(a0, z) ≤ #oc(a0, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
#oc(a0, x) < #oc(a0, y) + #oc(a0, z), ∀ a. #oc(a, x) ≤ #oc(a, y) + #oc(a, z), ∀ a. #oc(a, y) + #oc(a, z) ≤ #oc(a, x) ⊦ #oc(a0, y) + #oc(a0, z) = 0

Rule argument

Additional information

Used simplifier rules: f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat