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Proof step 18 from proof less-mult-div-01 (previous step)

i * i1 = ((i * i1 + i3 * i1) + i2 * (i + i3)) + j0, 1 < i1 + i2, 0 < i2, 0 < i1, 0 ≤ j0, j0 < i1 + i2, 0 < j1, 0 < i, 0 < i + i3, 0 ≤ i3 ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2