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Proof step 9 from proof ocs-dups-ex (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
∀ a1. a1 ∈ x0 → #oc(a1, x0) = 1, a = a0 ∨ a ∈ x0 ⊦ dups(x0), a0 ∈ x0, a ≠ a0 ∧ #oc(a, x0) = 1, ¬ a ∈ x0 ∧ #oc(a, a0 ') = 1

Rule argument

Additional information

Used simplifier rules: ocappend: ⊦ #oc(a, x + y) = #oc(a, x) + #oc(a, y);%oc-zero: ⊦ #oc(a, x) = 0 ↔ ¬ a ∈ x;% ⊦ dups(a ' + x) ↔ dups(x) ∨ a ∈ x; (ws)% ⊦ a ∈ x + y ↔ a ∈ x ∨ a ∈ y; (s)%c: ⊦ m + n = n + m; from specification nat-basic%a: ⊦ (m + n0) + n1 = m + n0 + n1; from specification nat-basic% ⊦ a ∈ b ' ↔ a = b; (s)% ⊦ m + n = 1 ↔ m = 0 ∧ n = 1 ∨ n = 0 ∧ m = 1; (ws)%a: ⊦ (x + y) + z = x + y + z; from specification list% ⊦ a + x = a ' + x; (s)