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Proof step 3 from proof ocnodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 4:
n ≤ #oc(a, a0 ') + #oc(a, x0), ∀ n0, a1. n0 ≤ #oc(a1, x0) → n0 = 0 ∨ n0 = 1 ⊦ n = 0, n = 1, a0 ∈ x0, dups(x0)

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%rec: ⊦ dups(a ' + x) ↔ ¬ ¬(a ∈ x ∨ dups(x)); from specification list-dup%a: ⊦ (x + y) + z = x + y + z; from specification list%c: ⊦ a + x = a ' + x; from specification list