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Proof step 11 from proof div-twoodd-07 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 12:
i < 0, 0 < j ⊦ (2 * i - 1) / (2 * j) = i / j
Proof step 13:
¬(i + 1 < 0 ∧ 0 < ~ j), j * 2 < 0, i < 0, j < 0 ⊦ ~((2 * (i + 1) - 1) / (2 * ~ j)) = (i + 1) / j

Rule argument

spec: int-div inst: name: div-twoodd-03 seq: i < 0, 0 < j ⊦ (2 * i - 1) / (2 * j) = i / j subst: [j, i] → [~ j, i + 1] flags: rewrite, current

Additional information

Used simplifier rules: neg: j ≠ 0 ⊦ i / ~ j = ~(i / j);%numls-03: ⊦ i + (* j) < (* j0) ↔ i < j0 - j; 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%neg-02: ⊦ ~ ~ i = i; from specification int-abs Used terms: ~ j, i + 1