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Proof step 3 from proof div-add-succ (previous step)

i + j - 1 = (i - 1) + j, 0 < j, 1 < j, 0 < i ⊦ i % j = 0, (i + j - 1) / j = i / j + 1

User applied rule insert equation

And obtained the following new goals

Proof step 4:
0 < j, 1 < j, 0 < i ⊦ i % j = 0, ((i - 1) + j) / j = i / j + 1

Rule argument

pos: left 1 flags: discard

Additional information