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Proof step 2 from proof div-twoodd-06 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-div inst: name: div-twoodd-02 seq: 0 ≤ i, j < 0 ⊦ (2 * i + 1) / (2 * j) = i / j subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used simplifier rules: mult-02: ⊦ 0 ≤ i0 ∧ i0 < j ∧ 0 ≤ i ∧ 0 < j0 → (i * j + i0) / (j0 * j) = i / j0;%zero-gels-04: ⊦ i ≤ 0 ∧ 0 ≤ j → i ≤ j; from specification int-abs%lst-01: ⊦ ¬ i < j → (i ≠ j ↔ j < i); from specification int-basic1%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%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: j, i