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Proof step 2 from proof div-mult-neg (previous step)

abs(i0) < j, 0 < j, i ≤ 0, i < j, i0 ≤ 0, i0 < j ⊦ (j * i + i0) / j = i

User applied rule cut formula

And obtained the following new goals

Proof step 3:
j * i + i0 = ~(j * ~ i + ~ i0), abs(i0) < j, 0 < j, i ≤ 0, i < j, i0 ≤ 0, i0 < j ⊦ (j * i + i0) / j = i

Rule argument

fma: j * i + i0 = ~(j * ~ i + ~ i0)

Additional information

Used simplifier rules: minus-left: ⊦ ~ i * j = ~(i * j); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%neg-11: ⊦ ~ i + ~ j = ~(i + j); from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%neg-02: ⊦ ~ ~ i = i; from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2 Used formulas: j * i + i0 = ~(j * ~ i + ~ i0)