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Proof step 19 from proof modmod-add-if (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information

Used simplifier rules: mod-ls-same-01: j ≠ 0 ⊦ i % j < i ↔ 0 < i ∧ (abs(j) = 1 ∨ abs(j) ≤ i);%mod-one: ⊦ i % 1 = 0;%mod-zero-09: ⊦ 0 < i ∧ i < j → i % j = i;%lst-03: ⊦ i + j < i0 ∧ ¬ j < j0 → i + j0 < i0; from specification int-basic2%forward-le: ⊦ i ≤ j ∧ 0 < i → 0 < j; (forward) from specification int-abs%lenls: ⊦ i ≤ j → ¬ j < i; from specification int-abs%add-zero-rev: ⊦ 0 + i = i; from specification int-basic2%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; from specification int-abs%sub-zero-01: ⊦ i - 0 = i; from specification int-abs%ls-01: ⊦ i + j < i + i0 ↔ j < i0; from specification int-basic2%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%succ-01: ⊦ i +1 = i + 1; from specification int-basic2%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs