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Proof step 1 from proof ocnodups (conclusion)

⊦ n ≤ #oc(a, x) ∧ ¬ dups(x) → n ≤ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 2:
n ≤ #oc(a, x) ⊦ n = 0, n = 1, dups(x)

Rule argument

Additional information

Used simplifier rules: one: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); from specification nat-basic1%lels-01: ⊦ ¬ m ≤ n ↔ n < m; from specification nat