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Proof step 5 from proof four-square-odd-ls (previous step)

0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, abs(i3) ≤ (2 * j + 1) / 2, abs(i4) ≤ (2 * j + 1) / 2, abs(i5) ≤ (2 * j + 1) / 2, abs(i6) ≤ (2 * j + 1) / 2, (2 * j + 1)^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j ⊦ j = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 6:
⊦ i % j = 0 ∧ 0 ≤ i ∧ 0 ≤ i0 ∧ 0 < j → (i + i0) / j = i / j + i0 / j
Proof step 7:
(2 * j) % 2 = 0 ∧ 0 ≤ 2 * j ∧ 0 ≤ 1 ∧ 0 < 2, 0 < i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, abs(i3) ≤ (2 * j) / 2 + 1 / 2, abs(i4) ≤ (2 * j) / 2 + 1 / 2, abs(i5) ≤ (2 * j) / 2 + 1 / 2, abs(i6) ≤ (2 * j) / 2 + 1 / 2, (2 * j + 1)^2 ≤ i3 ^2 + i4 ^2 + i5 ^2 + i6 ^2, 0 < j ⊦ j = 0

Rule argument

spec: int-div inst: 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] → [2, 1, 2 * j] flags: rewrite

Additional information

Used simplifier rules: mod-two-02: ⊦ (i * 2) % 2 = 0; from specification int-div%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%zero-ge: ⊦ 0 < i → (0 ≤ i * j ↔ 0 ≤ j); from specification int-mult%lels-02: ⊦ i ≠ j → (i ≤ j ↔ i < j); from specification int-abs Used terms: 2, 1, 2 * j