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Proof step 1 from proof div-twoodd-02 (conclusion)

0 ≤ i, j < 0 ⊦ (2 * i + 1) / (2 * j) = i / j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ j < 0 → i / (* j) = ~(i / ~ j)
Proof step 3:
2 * j < 0, 0 ≤ i, j < 0 ⊦ ~((2 * i + 1) / ~(2 * j)) = i / j

Rule argument

spec: int-div inst: name: num-neg-01 seq: ⊦ j < 0 → i / (* j) = ~(i / ~ j) subst: [j, i] → [2 * j, 2 * i + 1] flags: rewrite, current

Additional information

Used simplifier rules: zero-ls: ⊦ 0 < i → (i * j < 0 ↔ j < 0); from specification int-mult%zero-gels-01: ⊦ i < 0 ∧ 0 ≤ j → i < j; from specification int-abs%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult Used terms: 2 * j, 2 * i + 1