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Proof step 14 from proof div-mult-three-nls (previous step)

2 < j, i0 < j, j0 < 3, j * j * j < 3 * (j + 1), 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j ⊦ j = 2, j = 3

User applied rule cut formula

And obtained the following new goals

Proof step 15:
0 < j, j * j * j < 3 * (j + 1), j0 < 3, i0 < j, 2 < j, 3 * (j + 1) < j * (j + 1), 0 ≤ j, 0 ≤ i0, 0 ≤ j0 ⊦ j = 3, j = 2
Proof step 16:
0 < j, j * j * j < 3 * (j + 1), j0 < 3, i0 < j, 2 < j, 0 ≤ j, 0 ≤ i0, 0 ≤ j0 ⊦ j = 3, j = 2, 3 * (j + 1) < j * (j + 1)

Rule argument

fma: 3 * (j + 1) < j * (j + 1)

Additional information

Used formulas: 3 * (j + 1) < j * (j + 1)