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Proof step 12 from proof mult-def-02 (previous step)

gcd(i1, j1) = 1, 0 < j1, 0 < i, 1 < i, 0 < i3, ∀ j2, j1, i4, i3. i→n(i3) + i→n(i4) < i→n(i1) + i→n(i * i0) → 0 < j1 ∧ 0 < j2 → rational(i3, j1) * rational(i4, j2) = rational(i3 * i4, j1 * j2) ⊦ i→n(i1) + i→n(i0) < i→n(i1) + i→n(i * i0), rational(i1, j1) * rational(i0, i3) = rational(i1 * i * i0, j1 * i * i3)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 13:
gcd(i1, j1) = 1, 0 < i3, 1 < i, 0 < i, 0 < j1, ∀ j2, j1, i4, i3. i→n(i3) + i→n(i4) < i→n(i1) + i→n(i * 0) → 0 < j1 ∧ 0 < j2 → rational(i3, j1) * rational(i4, j2) = rational(i3 * i4, j1 * j2) ⊦ rational(i1, j1) * rational(0, i3) = rational(0, j1 * i * i3)

Rule argument

Additional information

Used simplifier rules: ⊦ i * 0 = 0;% ⊦ m + n = n + m;% ⊦ (m + n0) + n1 = m + n0 + n1;% ⊦ 1 < i → (i * abs(j) ≤ abs(j) ↔ j = 0);% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i * j = j * i;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ 1 < i → abs(i) = i;% ⊦ abs(i * j) = abs(i) * abs(j);% ⊦ i→n(i) < i→n(j) ↔ abs(i) < abs(j);% ⊦ m + n < m + n0 ↔ n < n0;