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Proof step 3 from proof disj-rec (
previous step
)
⊦ disj(asar, n, m) ↔ (∀ n0, n1. n0 < m ∧ n1 < m ∧ n0 ≠ n1 → asar[n + n0] ∩ asar[n + n1] = ∅)
Lemma
disj