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Proof step 3 from proof ocnodups (previous step)

n ≤ #oc(a, x) ⊦ n = 0, n = 1, dups(x)

User applied rule structural induction

And obtained the following new goals

Proof step 4:
n ≤ #oc(a, a0 + x0), ∀ n, a. n ≤ #oc(a, x0) → n = 0 ∨ n = 1 ∨ dups(x0) ⊦ n = 1, n = 0, dups(a0 + x0)

Rule argument

indvar: x

Additional information

Used simplifier rules: oc-e: ⊦ #oc(a, []) = 0;%list freely generated by [], +;;% ⊦ m ≤ 0 ↔ m = 0; (s)