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Proof step 6 from proof ocnodups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
n ≤ #oc(a, x0), ∀ n0, a1. n0 ≤ #oc(a1, x0) → n0 = 0 ∨ n0 = 1 ⊦ a0 = a, n = 0, n = 1, dups(x0), a0 ∈ x0

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ n + 0 = n; (s)