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Proof step 4 from proof div-twoodd-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 5:
~(2 * i - 1) = 2 * ~ i + 1, 2 * i - 1 < 0, 0 < j, i < 0 ⊦ 2 * j = 0, ~(~(2 * i - 1) / (2 * j)) = i / j

Rule argument

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

Additional information

Used simplifier rules: ⊦ ~(i - 1) = ~ i + 1; (s)% ⊦ i + ~ j = i - j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ~ i * j = ~(i * j); (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i - j = i0 ↔ i = i0 + j; (s)% ⊦ i + (j - i) = j; (s) Used formulas: ~(2 * i - 1) = 2 * ~ i + 1