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Proof step 5 from proof mod-add1-negpos-pos (previous step)

i = i1 * j + j0, j0 = 0, 0 < j, i < 0, abs(j0) < j, 0 ≤ i0, 0 ≤ i + i0, j0 ≤ 0, i1 ≤ 0 ⊦ (i + i0) % j = i0 % j

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
i1 < 0, 0 < j, 0 ≤ i0, 0 ≤ i1 * j + i0 ⊦ (i1 * j + i0) % j = i0 % j

Rule argument

Additional information

Used simplifier rules: ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)%forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs%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% ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; (s)%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ 0 < i → (i * j < 0 ↔ j < 0); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + 0 = i; (s)