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Proof step 2 from proof mod-twoodd-03 (previous step)

i < j, i < 0, 0 < j, i + 1 < j ⊦ (i + 1) % j * 2 = (2 * i + 1) % (2 * j) + 1

User applied rule cut formula

And obtained the following new goals

Proof step 3:
2 * i + 1 = ~(2 * (~ i - 1) + 1), i < j, i < 0, 0 < j, i + 1 < j ⊦ (i + 1) % j * 2 = (2 * i + 1) % (2 * j) + 1

Rule argument

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

Additional information

Used simplifier rules: subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%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-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-sub: ⊦ ~(i - j) = j - i; from specification int-abs%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%eq-dist: ⊦ (i + j) * i0 = j0 + j * i0 ↔ i * i0 = j0; from specification int-mult Used formulas: 2 * i + 1 = ~(2 * (~ i - 1) + 1)