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Proof step 11 from proof div-add-succ (previous step)

i0 * j + j0 - 1 = (j0 - 1) + i0 * j, j0 < j, 0 < j, 0 ≤ j, 0 ≤ i0, 0 ≤ i0 * j + j0, 0 < i0 * j + j0, 0 < j0, 1 < j ⊦ j0 = 0, (i0 * j + j0 - 1) / j = i0

User applied rule insert equation

And obtained the following new goals

Rule argument

pos: left 1 flags: discard

Additional information