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Proof step 10 from proof div-mult-two-nls-01 (previous step)

2 * i0 * j + 2 * j0 = i1 * j + j1, j1 < j, j0 < j, 1 + 2 * i0 < i1, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j1, 0 ≤ i1 ⊦

User applied rule cut formula

And obtained the following new goals

Proof step 11:
2 * i0 * j = (2 * i0) * j, 2 * i0 * j + 2 * j0 = i1 * j + j1, j1 < j, j0 < j, 1 + 2 * i0 < i1, 0 < j, 0 ≤ j0, 0 ≤ i0, 0 ≤ j1, 0 ≤ i1 ⊦

Rule argument

fma: 2 * i0 * j = (2 * i0) * j

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 Used formulas: 2 * i0 * j = (2 * i0) * j