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Proof step 2 from proof ocnodups (previous step)

n ≤ #oc(a, x) ⊦ dups(x), n ≤ 1

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 3:
n ≤ #oc(a, x) ⊦ n = 0, n = 1, dups(x)

Rule argument

Additional information

Used simplifier rules: ⊦ 1 < m ↔ ¬(m = 0 ∨ m = 1); (s)% ⊦ m ≤ n ↔ ¬ n < m; (ss)