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Proof step 1 from proof mod-mult-neg (conclusion)

⊦ 0 < j ∧ i ≤ 0 ∧ i0 ≤ 0 ∧ abs(i0) < j → (j * i + i0) % j = i0

User applied rule cut formula

And obtained the following new goals

Proof step 2:
j * i + i0 = ~(j * ~ i + ~ i0) ⊦ 0 < j ∧ i ≤ 0 ∧ i0 ≤ 0 ∧ abs(i0) < j → (j * i + i0) % j = i0

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)