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Proof step 1 from proof oc-dups (conclusion)

⊦ n < #oc(a, x) ∧ ¬ dups(x) → #oc(a, x) = 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n < #oc(a, x), a ∈ x ⊦ #oc(a, x) = 1, dups(x)

Rule argument

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%f: ⊦ m < n → n ≠ 0; (forward) from specification nat-basic1