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Proof step 2 from proof gcd-div-rest (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 3:
i = i1 * gcd(i, j) + j0, abs(j0) < gcd(i, j) ⊦ j0 = 0, i = i1 * gcd(i, j) ∧ j0 = 0

Rule argument

fma: j0 = 0

Additional information

Used simplifier rules: add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: j0 = 0