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Proof step 3 from proof div-twoodd-05 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
i < 0, 0 < j ⊦ (2 * i - 1) / (2 * j) = i / j
Proof step 5:
¬(i + 1 < 0 ∧ 0 < j), i < 0, 0 < j ⊦ (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 terms: j, i + 1