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Proof step 5 from proof oc-dups (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
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)

Rule argument

Additional information

Used simplifier rules: one-01: ⊦ a ≠ b → #oc(a, b ') = 0;%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;%c: ⊦ m + n = n + m; from specification nat-basic2%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic2%add-base: ⊦ n + 0 = n; from specification nat-basic2