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Proof step 17 from proof gcd-rec (previous step)

abs(i) = i0 * abs(j) + j0, j0 < abs(j), 0 ≤ j0, 0 ≤ i0, gcd(abs(i), abs(j)) < gcd(abs(j), j0) ⊦ gcd(abs(i), abs(j)) = gcd(abs(j), j0), j = 0, i = 0, gcd(abs(j), j0) | abs(i)

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