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Proof step 4 from proof perm-dups (previous step)

dups(y), x ⊆m y, ∀ a. #oc(a, y) ≤ #oc(a, x) ⊦ dups(x)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 5:
⊦ dups(x) ↔ (∃ a, x0, y, z. x = x0 + (a + (y + (a + z))))
Proof step 6:
∃ a, x0, y0, z. y = x0 + (a + (y0 + (a + z))), x ⊆m y, ∀ a. #oc(a, y) ≤ #oc(a, x) ⊦ dups(x)

Rule argument

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

Additional information

Used terms: y