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Proof step 6 from proof modmod-add-if3 (previous step)

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

Rule argument

pos: right 1

Additional information

Used simplifier rules: zero-gs: ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2