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Proof step 14 from proof modmod-add-if3 (previous step)

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

Additional information

Used simplifier rules: zero-ge-01: ⊦ i < 0 ∧ j < 0 → ¬ 0 ≤ i + j; 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