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Proof step 3 from proof div-mult-pos-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div-add-pos seq: ⊦ i0 % j = 0 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j subst: [j, i0, i] → [j, i0 * j, i] flags: <all paths>

Additional information

Used simplifier rules: mod-mult-04: j ≠ 0 ⊦ (i * j) % j = 0;%div-mult-03: j ≠ 0 ⊦ (i * j) / j = i;% ⊦ i * 0 = 0; (s)%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ i + 0 = i; (s)% ⊦ i = i + j ↔ j = 0; (s)% ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; (s)% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s) Used terms: j, i0 * j, i