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Proof step 24 from proof equality-pos (previous step)

0 < i0 ∧ 0 < i ∧ 0 ≤ j, i * j0 = i0 * j, 0 < j, 0 < j0, 0 < i, 0 < i0 ⊦ j0 = 0, j = 0, i = 0, i0 = 0, i = gcd(i * i0, j * i0) / gcd(i0, j0) ∧ j = (j0 * gcd(j, i)) / gcd(j0, i0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ 0 < i1 ∧ 0 < i ∧ 0 ≤ j → gcd(i * i1, j * i1) = gcd(i, j) * i1
Proof step 26:
0 < j0 ∧ 0 < j ∧ 0 ≤ i, 0 < i0 ∧ 0 < i ∧ 0 ≤ j, i * j0 = i0 * j, 0 < j, 0 < j0, 0 < i, 0 < i0 ⊦ j0 = 0, j = 0, i = 0, i0 = 0, i = gcd(i * i0, j * i0) / gcd(i0, j0) ∧ j = gcd(j * j0, i * j0) / gcd(j0, i0)

Rule argument

seq: ⊦ 0 < i1 ∧ 0 < i ∧ 0 ≤ j → gcd(i * i1, j * i1) = gcd(i, j) * i1 subst: [i1, j, i] → [j0, i, j] rotate: T spec: int-gcd inst: name: gcd-mult-pos

Additional information

Used simplifier rules: ⊦ i ≠ j → (i ≤ j ↔ i < j);% ⊦ gcd(i * i1, j * i1) = gcd(i, j) * abs(i1);% ⊦ 0 < j → abs(j) = j;% ⊦ i * j * j0 = (i * j) * j0;% ⊦ ¬ i < j ↔ j ≤ i;% ⊦ i < j → ¬ j < i;% ⊦ ¬ i ≤ j ↔ j < i;% ⊦ i * j = j * i; Used terms: j0, i, j