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Proof step 15 from proof div-mult-two-nls-01 (previous step)

2 * (j0 + i0 * j) = i1 * j + j1, j1 < 2 * j0, j ≤ 2 * j0, 0 ≤ i1, 0 ≤ j1, 0 ≤ i0, 0 ≤ j0, 1 + 2 * i0 < i1, j0 < j, j1 < j, 1 < j, 0 < j, 0 < j0 ⊦

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: 2 * (j0 + i0 * j) = (2 * i0 + 1) * j + (2 * j0 - j)

Additional information

Used simplifier rules: sub-add-01: ⊦ i + (j - i0) = i + j - i0; 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%sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%eq-succ: ⊦ i * (i0 + 1) + j = i + j0 ↔ i * i0 + j = j0; from specification int-mult%dist-rmerge: ⊦ i * j + i * j0 = i * (j + 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%eq-pos: ⊦ 0 < i → (i * j = i * j0 ↔ j = j0); from specification int-mult Used formulas: 2 * (j0 + i0 * j) = (2 * i0 + 1) * j + (2 * j0 - j)