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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), 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) | (i0 * i3) * gcd(i, j), i = 0

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

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