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Proof step 6 from proof div-mult-ls (previous step)

i * (i1 * j + j0) = i * i1 * j + i * j0, j0 < j, (i * (i1 * j + j0)) / j < i * i1, 0 ≤ i, 0 < j, 0 ≤ j0, 0 ≤ i1 ⊦

User applied rule insert equation

And obtained the following new goals

Proof step 7:
j0 < j, (i * i1 * j + i * j0) / j < i * i1, 0 ≤ i, 0 < j, 0 ≤ j0, 0 ≤ i1 ⊦

Rule argument

pos: left 1 flags: discard

Additional information