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Proof step 28 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 < j, 0 < i, 0 ≤ i1, 0 ≤ j, 0 ≤ i3 ⊦ i = 0, i0 * gcd(i, j) | i

User applied rule insert equation

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: div-mult: ⊦ i | i * j;%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult