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Proof step 2 from proof modmod-add-if (previous step)

0 < j, 0 ≤ i ⊦ (i + i0 % j) % j = (0 ≤ i0 ⊃ (i + i0) % j;(0 ≤ i + i0 ⊃ (i + i0) % j;(i + i0 % j < 0 ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j;0))))

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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