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Proof step 14 from proof diff (previous step)

∀ a. #oc(a, y0) ≤ #oc(a, y), ∀ a. #oc(a, y) ≤ #oc(a, y0), ∀ a. #oc(a, x0) ≤ #oc(a, x), ∀ a. #oc(a, x) ≤ #oc(a, x0) ⊦ a0 ∈ y0, #oc(a0, x0 \ y0) ≤ #oc(a0, x \ y)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
#oc(a0, x \ y) < #oc(a0, x0), a0 ∈ x0, ∀ 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) ⊦ a0 ∈ y0

Rule argument

Additional information

Used simplifier rules: oc-diff-01: ⊦ ¬ a ∈ y → #oc(a, x \ y) = #oc(a, x);%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