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Proof step 20 from proof gcd-div-gcd-pos (previous step)

j = (i0 * i1) * gcd(j, i), i = (i0 * i3) * gcd(i, j), 0 ≤ j, 0 < i, 0 < j, 0 < i0, 1 < i0, 0 ≤ i, 0 < gcd(i, j), 0 ≤ i3, 0 < i3, 0 ≤ i1, 0 < i1, 0 < gcd(j, i) ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 21:
⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j)
Proof step 22:
i0 * gcd(i, j) | i ∧ i0 * gcd(i, j) | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 * gcd(i, j) ≤ gcd(i, j), j = (i0 * i1) * gcd(j, i), i = (i0 * i3) * gcd(i, j), 0 ≤ j, 0 < i, 0 < j, 0 < i0, 1 < i0, 0 ≤ i, 0 < gcd(i, j), 0 ≤ i3, 0 < i3, 0 ≤ i1, 0 < i1, 0 < gcd(j, i) ⊦

Rule argument

spec: int-gcd inst: name: gcd-max seq: ⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j) subst: [i0, i, j] → [i0 * gcd(i, j), i, j] flags: current, []

Additional information

Used terms: i0 * gcd(i, j), i, j