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Theorem subarray-eq-elim

n + m ≤ # ar, n0 = # ar ⊦ subarray(ar, n, m) = ar1 ↔ (∃ ar0, ar2. ar = ar0 + ar1 + ar2 ∧ # ar0 = n ∧ # ar2 = n0 - n + m)

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