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Proof step 10 from proof del-02 (previous step)

a ∈ x, a ∈ y, ∀ a0. #oc(a0, x -1l a) ≤ #oc(a0, y -1l a), #oc(a1, y -1l a) ≤ #oc(a1, x -1l a) ⊦ #oc(a1, y) ≤ #oc(a1, x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 11:
#oc(a1, x) < #oc(a1, y), #oc(a1, y -1l a) ≤ #oc(a1, x -1l a), a ∈ y, a ∈ x, a1 ∈ y, ∀ a0. #oc(a0, x -1l a) ≤ #oc(a0, y -1l a) ⊦

Rule argument

Additional information

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