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Proof step 29 from proof equality-pos (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 30:
j ≠ 0 ⊦ (j * i) / j = i
Proof step 31:
gcd(j0, i0) ≠ 0, 0 < j ∧ 0 < j0 ∧ 0 ≤ i0, 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 = j

Rule argument

seq: j ≠ 0 ⊦ (j * i) / j = i subst: [i, j] → [j, gcd(j0, i0)] rotate: F spec: int-div inst: name: div-mult

Additional information

Used simplifier rules: ⊦ gcd(i, j) = 0 ↔ i = 0 ∧ j = 0; Used terms: j, gcd(j0, i0)