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Proof step 5 from proof div-mult-three-nls (previous step)

j * j * j = i0 * 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 6:
i0 * 3 + j0 < j * 3 + j0, j * j * j = i0 * 3 + j0, i0 < j, 2 ≤ j, 0 < j, j0 < 3, 0 ≤ j, 0 ≤ j0, 0 ≤ i0 ⊦

Rule argument

fma: i0 * 3 + j0 < j * 3 + j0

Additional information

Used simplifier rules: ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%ls-pos: ⊦ 0 < i → (i * j < i * j0 ↔ j < j0); from specification int-mult%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%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 Used formulas: i0 * 3 + j0 < j * 3 + j0