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Proof step 2 from proof modmod-add2-negpos-pos-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And closed the goal!

Rule argument

Additional information

Used simplifier rules: modmod-add-zero: 0 < j ⊦ (i + i0 % j) % j = 0 → (i + i0) % j = 0;% ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ i < j ↔ ¬ j ≤ i; (ss)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2