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Proof step 7 from proof div-mult-three-nls (previous step)

j * j * j < j * 3 + j0, i0 < j, 2 ≤ j, 0 < j, j0 < 3, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 8:
j * j * j < j * 3 + j0, i0 < j, 2 ≤ j, 0 < j, j0 < 3, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦ j = 2

Rule argument

fma: j = 2

Additional information

Used simplifier rules: numls-02: ⊦ (* j) < i0 + (* j0) ↔ j - j0 < i0; from specification int-abs%num-between: i + 1 = i0 ⊦ j < (* i0) → ¬ (* i) < j; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: j = 2