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Proof step 27 from proof gcd-mult-pos (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 28:
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(i * i1, j * i1) = gcd(i, j) * i1

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i < j ↔ ¬ j ≤ i; (ss)