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Proof step 11 from proof gcd-rec (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 12:
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(j) ∧ gcd(abs(j), j0) | abs(i)

Rule argument

Additional information

Used simplifier rules: gcd-zero-04: ⊦ gcd(i, 0) = abs(i);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%gcd-nls-01: ⊦ i ≠ 0 → ¬ abs(i) < gcd(abs(i), j);%abs-01: ⊦ abs(abs(i)) = abs(i); from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs