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Proof step 20 from proof modmod-add-if3 (previous step)

0 ≤ i + i0, i < j, i < 0, 0 < j, i + 1 < j, 0 ≤ i0 ⊦ (0 ≤ i + i0 % j ⊃ (i + i0) % j;((i + i0 % j) % j ≠ 0 ⊃ (i + i0) % j - j;0)) = (i + i0 % j < 0 ∧ (i + i0 % j) % 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: modmod-add-zero: 0 < j ⊦ (i + i0 % j) % j = 0 → (i + i0) % j = 0;%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → 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%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs