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Proof step 2 from proof gcd-mult-pos (previous step)

0 < i1, 0 < i, 0 ≤ j ⊦ gcd(i * i1, j * i1) = gcd(i, j) * i1

User applied rule induction

And obtained the following new goals

Proof step 3:
0 < i1 ∧ 0 < i ∧ 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 * i1) = gcd(i, j) * i1

Rule argument

indvar: i→n(j * i1 + i * i1) wfpred: <

Additional information

Used simplifier rules: <; (wellfounded) Used terms: i→n(j * i1 + i * i1)