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Proof step 9 from proof gcdhex (previous step)

j ≠ 0 → gcd(i, j) = gcd(abs(j), abs(i) % abs(j)), 0 ≤ i, j ≤ i, 0 < i, 0 < j, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
gcd(i, j) = gcd(j, i % j), 0 ≤ i, j ≤ i, 0 < i, 0 < j, ∀ i0, j. i→n(i0) < i→n(i) → 0 ≤ i0 ∧ 0 ≤ j ∧ j ≤ i0 → (∃ i1, j0. gcd(i0, j) = i0 * i1 + j * j0) ⊦ j = 0, ∃ i0, j0. gcd(i, j) = i * i0 + j * j0

Rule argument

Additional information

Used simplifier rules: abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs