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Proof step 8 from proof modmod-add-if3 (previous step)

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

Proof step 9:
0 < j, i0 < 0, 0 ≤ i ⊦ (i + i0 % j < 0 ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j;0)) = (0 ≤ i + i0 % j ∧ i + i0 < 0 ∧ (i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j;(i + i0) % j), 0 ≤ i + i0

Rule argument

pos: right 1

Additional information

Used simplifier rules: ⊦ i ≤ j → ¬ j < i; (s)