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

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

User applied rule cut formula

And closed the goal!

Rule argument

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

Additional information

Used simplifier rules: associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i < j ∧ ¬ i0 < j → ¬ i0 < i; (s)% ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); (s)% ⊦ i < j + 1 ↔ i ≤ j; (s)% ⊦ 0 < i → (i * j ≤ i ↔ j ≤ 1); (s)% ⊦ 0 < i → (i ≤ 1 ↔ i = 1); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i = 1 → i * j = j; (s)% ⊦ (* i1) * j < j ↔ 0 < (1 - i1) * j; (s)% ⊦ i < 0 → (0 < i * j ↔ j < 0); (s)% ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0; (s) Used formulas: j * j * j < j * (j + 1)