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Proof step 27 from proof gcd-div-gcd-pos (previous step)

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

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: [[2,1,2]]

Additional information