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Proof step 2 from proof ocone (previous step)

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;% ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)% ⊦ m < 1 ↔ m = 0; (s)% ⊦ m ≠ n → (m ≤ n ↔ m < n); (s)