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Proof step 12 from proof nodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
#oc(a, y) = 1, #oc(a, x) = 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;% ⊦ 0 < n ↔ n ≠ 0; (s)%lf-02: ⊦ m < n ∧ m = n0 → n0 < n; (forward) from specification nat-basic% ⊦ m < 1 ↔ m = 0; (s)%lf-01: ⊦ m < n ∧ n = n0 → m < n0; (forward) from specification nat-basic% ⊦ m < n → n ≠ 0; (s)%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic% ⊦ m ≤ n ↔ ¬ n < m; (ss)