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Proof step 6 from proof oc-dups (previous step)

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

Heuristic quantifier closing applied rule all left

And closed the goal!

Rule argument

pos: left 3 terms: n, a

Additional information

Used simplifier rules: oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%lels: ⊦ ¬ m < n ↔ n ≤ m; from specification nat%fle-01: ⊦ m ≤ n ∧ n < n0 → m < n0; (forward) from specification nat%l-01: ⊦ m < m + n ↔ n ≠ 0; from specification nat-basic2%in-17: ⊦ a ∈ b ' ↔ a = b; from specification list%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%c: ⊦ m + n = n + m; from specification nat-basic2 Used terms: n, a