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Proof step 9 from proof div-mult-three-nls (previous step)

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

User applied rule cut formula

And obtained the following new goals

Rule argument

fma: j * j * j < j * 3 + 3

Additional information

Used simplifier rules: succ-rmerge: ⊦ i * j + i = i * (j + 1); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%forward-lels: ⊦ i ≤ j ∧ j < j0 → i < j0; (forward) from specification int-abs%forward-pos: ⊦ i < j ∧ 0 < i → 0 < j ∧ 1 < j; (forward) from specification int-basic1%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used formulas: j * j * j < j * 3 + 3