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Proof step 15 from proof gcd-mult-pos (previous step)

i ≠ 0, i < j, 0 < i1, 0 < i, 0 ≤ j, 1 < j, 0 < j, ∀ j0, i0, i2. i→n(j0 * i0 + i2 * i0) < i→n(j * i1 + i * i1) → 0 < i0 ∧ 0 < i2 ∧ 0 ≤ j0 → gcd(i2 * i0, j0 * i0) = gcd(i2, j0) * i0 ⊦ gcd(i * i1, j % i * i1) = gcd(abs(i), abs(j) % abs(i)) * i1, i1 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
i < j, 0 < i1, 0 < i, 0 ≤ j, 1 < j, 0 < j, ∀ j0, i0, i2. i→n(j0 * i0 + i2 * i0) < i→n(j * i1 + i * i1) → 0 < i0 ∧ 0 < i2 ∧ 0 ≤ j0 → gcd(i2 * i0, j0 * i0) = gcd(i2, j0) * i0 ⊦ i1 = 0, i = 0, gcd(i * i1, j % i * i1) = gcd(i, j % i) * i1

Rule argument

Additional information

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