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Proof step 2 from proof oc-dups (previous step)

n < #oc(a, x), a ∈ x ⊦ #oc(a, x) = 1, dups(x)

User applied rule structural induction

And obtained the following new goals

Proof step 3:
∀ n, a. n < #oc(a, x0) ∧ a ∈ x0 → #oc(a, x0) = 1 ∨ dups(x0), n < #oc(a, a0 + x0), a ∈ a0 + x0 ⊦ #oc(a, a0 + x0) = 1, dups(a0 + x0)

Rule argument

indvar: x

Additional information

Used simplifier rules: oc-e: ⊦ #oc(a, []) = 0;%list freely generated by [], +;;%zero-01: ⊦ ¬ m < 0; from specification nat-basic1