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Proof step 13 from proof sub-del-any (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 14:
#oc(a0, y) < #oc(a0, x), a0 ∈ x, #oc(a, x) ≤ #oc(a, y), ∀ a. #oc(a, x) ≤ #oc(a, y) ⊦

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)