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Proof step 1 from proof div-twoodd-03 (conclusion)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

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

Additional information

Used simplifier rules: nummult-ls: ⊦ 0 < i ∧ 0 ≤ j → ((* i) * i0 < (* j) ↔ i0 < (i + j - 1) / i);%zero-neq-01: ⊦ 0 < i → (i * j = 0 ↔ j = 0); from specification int-mult%associativity: ⊦ i * j * j0 = (i * j) * j0; from specification int-mult%commutativity: ⊦ i * j = j * i; from specification int-mult%sub-ls: ⊦ i - j < i0 ↔ i < j + i0; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%lsone: ⊦ i < 1 ↔ i ≤ 0; from specification int-abs%zero-gels-03: ⊦ i < 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lf-02: ⊦ i < j ∧ j = i0 → i < i0; (forward) from specification int-basic1%forward-le-05: ⊦ i < j ∧ 0 ≤ i → 0 < j; (forward) from specification int-abs%lf-01: ⊦ i < j ∧ j < i0 → i +1 < i0; (forward) from specification int-basic1%trans: ⊦ i < j ∧ j < j0 → i < j0; (forward) from specification int-basic1%forward-le-02: ⊦ j < i ∧ i ≤ 0 → j < 0; (forward) from specification int-abs Used terms: 2 * j, 2 * i - 1