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Proof step 33 from proof gcd-mult-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 34:
0 < i1, 0 < i, 0 < j, j ≤ i, 0 ≤ i, ∀ j0, i0, i2. i→n(j0 * i0 + i2 * i0) < i→n(j * i1 + i * i1) → 0 < i2 ∧ 0 < i0 ∧ 0 ≤ j0 → gcd(i2 * i0, j0 * i0) = gcd(i2, j0) * i0 ⊦ gcd(j * i1, i % j * i1) = gcd(j, i % j) * i1, i1 = 0, j = 0

Rule argument

Additional information

Used simplifier rules: ⊦ 0 < j → abs(j) = j; (s)