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Proof step 3 from proof gcd-div-rest (previous step)

i = i1 * gcd(i, j) + j0, abs(j0) < gcd(i, j) ⊦ j0 = 0, i = i1 * gcd(i, j) ∧ j0 = 0

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information