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Proof step 23 from proof gcd-div-gcd-pos (previous step)

i = (i0 * i3) * gcd(i, j), j = (i0 * i1) * gcd(j, i), 0 < gcd(j, i), 0 < gcd(i, j), 1 < i0, 0 < i0, 0 < i, 0 < j, 0 ≤ i1, 0 ≤ j, 0 ≤ i3, 0 ≤ i, i0 * gcd(i, j) | i ∧ i0 * gcd(i, j) | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 * gcd(i, j) ≤ gcd(i, j) ⊦

User applied rule case distinction

And obtained the following new goals

Proof step 24:
i = (i0 * i3) * gcd(i, j), j = (i0 * i1) * gcd(j, i), 0 < j, 0 < i, 0 < i0, 1 < i0, 0 < gcd(i, j), 0 < gcd(j, i), 0 ≤ i, 0 ≤ i3, 0 ≤ j, 0 ≤ i1 ⊦ i0 * gcd(i, j) | i ∧ i0 * gcd(i, j) | j ∧ (i ≠ 0 ∨ j ≠ 0)

Rule argument

pos: left 13

Additional information

Used simplifier rules: gcd-com: ⊦ gcd(i, j) = gcd(j, i);% ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1); (s)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i = 1 → i * j = j; (s)