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Proof step 12 from proof div-addthree-rew (previous step)

j1 < j, 0 < j, j0 < j, j2 < j, 0 ≤ i3, 0 ≤ j1, 0 ≤ j0, 0 ≤ i2, 0 ≤ j2, 0 ≤ i0 ⊦ (j * (i0 + i3 + i2) + j0 + j1 + j2) / j = i2 + i3 + i0 + (j0 + j1 + j2) / j

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: <toplevel> name: div-add1-pospos seq: ⊦ i % j = 0 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j subst: [j, i0, i] → [j, j0 + j1 + j2, j * (i0 + i3 + i2)] flags: <all paths>

Additional information

Used simplifier rules: div-mult-pos-03: ⊦ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + j * i0) / j = i / j + i0;%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%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2% ⊦ i + j = i + i0 ↔ j = i0; (s)%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%forward-lsle: ⊦ i < j ∧ j ≤ j0 → i < j0; (forward) from specification int-abs%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs% ⊦ i + j < j ↔ i < 0; (s)% ⊦ 0 ≤ i ∧ 0 ≤ j → ¬ i + j < 0; (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s) Used terms: j, j0 + j1 + j2, j * (i0 + i3 + i2)