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Proof step 3 from proof div-twoodd-03 (previous step)

2 * j ≠ 0 ∧ 2 * i - 1 < 0, i < 0, 0 < j ⊦ ~(~(2 * i - 1) / (2 * j)) = i / j

User applied rule cut formula

And obtained the following new goals

Proof step 4:
~(2 * i - 1) = 2 * ~ i + 1, 2 * j ≠ 0 ∧ 2 * i - 1 < 0, i < 0, 0 < j ⊦ ~(~(2 * i - 1) / (2 * j)) = i / j

Rule argument

fma: ~(2 * i - 1) = 2 * ~ i + 1

Additional information

Used simplifier rules: minus-02: ⊦ ~(i - 1) = ~ i + 1; from specification int-abs%neg-15: ⊦ 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%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%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%Sub-def: ⊦ i + (j - i) = j; from specification int-abs Used formulas: ~(2 * i - 1) = 2 * ~ i + 1