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Proof step 5 from proof app-01 (previous step)

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

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)