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Proof step 1 from proof div-twoodd-07 (conclusion)

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

User applied rule cut formula

And obtained the following new goals

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

Rule argument

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

Additional information

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