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Proof step 18 from proof modmod-add-if3 (previous step)

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

Heuristic if-then-else split applied rule case distinction

And closed the goal!

Rule argument

pos: right 1

Additional information

Used simplifier rules: sub-eq: ⊦ i - j = i0 ↔ i = i0 + j; from specification int-abs%eq: ⊦ i = i + j ↔ j = 0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs