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Proof step 6 from proof nodups (previous step)

∀ a. a ∈ x ↔ a ∈ y ⊦ ∀ a. #oc(a, x) ≤ #oc(a, y), dups(y), dups(x)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
#oc(a, x) = 1, #oc(a, y) = 0, ∀ a. a ∈ x ↔ a ∈ y ⊦ dups(y), dups(x)

Rule argument

Additional information

Used simplifier rules: oc-dups: ⊦ n < #oc(a, x) ∧ ¬ dups(x) → #oc(a, x) = 1;%z: ⊦ 0 < n ↔ n ≠ 0; from specification nat-basic1%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic1%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic1%f: ⊦ m < n → n ≠ 0; from specification nat-basic1%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat