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Proof step 10 from proof trans (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
a0 ∈ z, #oc(a0, z) ≤ #oc(a0, y), #oc(a0, x) < #oc(a0, z), #oc(a0, x) < #oc(a0, y), a0 ∈ y, ∀ a. #oc(a, z) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, z), ∀ a. #oc(a, y) ≤ #oc(a, x), ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%fle: ⊦ m < n ∧ n ≤ n0 → m < n0; (forward) from specification nat%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat