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Proof step 16 from proof modmod-add-if3 (previous step)

0 ≤ i0, i + 1 < j, 0 < j, i < 0, i < j ⊦ (i + i0 < 0 ⊃ (i + i0) % j;(0 ≤ i + i0 % j ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;0))) = (i + i0 % j < 0 ∧ 0 ≤ i + i0 ∧ (i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;(i + i0) % j)

Heuristic if-then-else split applied rule case distinction

And obtained the following new goals

Rule argument

pos: right 1

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