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Proof step 1 from proof ocone (conclusion)

⊦ a ∈ x → (#oc(a, x) ≤ 1 ↔ #oc(a, x) = 1)

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: in-05: ⊦ a ∈ x → #oc(a, x) ≠ 0;%one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat%one-01: ⊦ m < 1 ↔ m = 0; from specification nat-basic1%lels-02: ⊦ m ≠ n → (m ≤ n ↔ m < n); from specification nat