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Proof step 5 from proof mod-mult-pos-01 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 6:
0 < j, j0 < j, 0 ≤ i, 0 ≤ j0, 0 ≤ i1 ⊦ (j * (i1 + i) + j0) % j = j0

Rule argument

Additional information

Used simplifier rules: forward-le-01: ⊦ i ≤ j ∧ 0 ≤ i → 0 ≤ j; (forward) from specification int-abs% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 ≤ j1 ∧ j1 < j → (0 ≤ i0 * j + j1 ↔ 0 ≤ i0); (s)%add-c: ⊦ i + j = j + i; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%commutativity: ⊦ i * j = j * i; from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult% ⊦ i * j + i * j0 = i * (j + j0); (s)