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Proof step 1 from proof div-mult-num (conclusion)
j ≠ 0, j0 % j = 0 ⊦ (i * (* j0)) / (* j) = i * j0 / j
Special step normalize
And obtained the following new goals
Proof step 2
:
j ≠ 0, j0 % j = 0 ⊦ (i * j0) / j = i * j0 / j