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Proof step 9 from proof diff (previous step)

a ∈ y0, #oc(a, x0 \ y0) < #oc(a, x), a ∈ x, ∀ a. #oc(a, x) ≤ #oc(a, x0), ∀ a. #oc(a, x0) ≤ #oc(a, x), ∀ a. #oc(a, y) ≤ #oc(a, y0), ∀ a. #oc(a, y0) ≤ #oc(a, y) ⊦ a ∈ y

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
a ∈ y0, a ∈ x, ∀ a. #oc(a, x) ≤ #oc(a, x0), ∀ a. #oc(a, x0) ≤ #oc(a, x), ∀ a. #oc(a, y) ≤ #oc(a, y0), ∀ a. #oc(a, y0) ≤ #oc(a, y) ⊦ a ∈ y

Rule argument

Additional information

Used simplifier rules: oc-diff: ⊦ a ∈ y → #oc(a, x \ y) = 0;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%zero: ⊦ 0 < m ↔ m ≠ 0; from specification nat-basic1