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Proof step 6 from proof modmod-add-if (previous step)

i0 < j, i0 < 0, 0 < j, 0 ≤ i, i0 + 1 < j ⊦ (i + i0 % j) % 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

Proof step 7:
¬ 0 ≤ i + i0, i0 < j, i0 < 0, 0 < j, 0 ≤ i, i0 + 1 < j ⊦ (i + i0 % j) % j = (i + i0 % j < 0 ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j + j;0))

Rule argument

pos: right 1

Additional information

Used simplifier rules: modmod-add2-posneg-pos: ⊦ 0 ≤ i + i0 ∧ 0 ≤ i ∧ i0 < 0 ∧ 0 < j → (i + i0 % j) % j = (i + i0) % j;%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2