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Proof step 3 from proof ocs-dups-ex (previous step)

dups(x) ⊦ ∃ a. a ∈ x ∧ #oc(a, x) ≠ 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ dups(x) ↔ (∃ a, x0, y, z. x = x0 + (a + (y + (a + z))))
Proof step 5:
∃ a, x0, y, z. x = x0 + (a + (y + (a + z))) ⊦ ∃ a. a ∈ x ∧ #oc(a, x) ≠ 1

Rule argument

spec: list-dup inst: name: dups seq: ⊦ dups(x) ↔ (∃ a, x0, y, z. x = x0 + (a + (y + (a + z)))) subst: [x] → [x] flags: rewrite

Additional information

Used terms: x