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Proof step 16 from proof div-mult-two-nls-01 (previous step)

2 * (j0 + i0 * j) = (2 * i0 + 1) * j + (2 * j0 - j), 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 insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: discard

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