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Proof step 14 from proof perm-all (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 15:
#oc(a0, x) < #oc(a0, y), a0 ∈ y, ∀ a. a ∈ x ∨ a ∈ y → #oc(a, x) = #oc(a, y) ⊦ a ∈ y, a ∈ x

Rule argument

Additional information

Used simplifier rules: notin: ⊦ ¬ a ∈ x → #oc(a, x) = 0;%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