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Proof step 2 from proof div2one-01 (previous step)

i % 2 = 1 ⊦ (1 + i / 2) * 2 = i + 1

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
j ≠ 0 ⊦ i % j = i - i / j * j
Proof step 4:
2 ≠ 0, i - i / 2 * 2 = 1 ⊦ (1 + i / 2) * 2 = i + 1

Rule argument

spec: int-div inst: name: mod-rew seq: j ≠ 0 ⊦ i % j = i - i / j * j subst: [j, i] → [2, i] flags: rewrite, current

Additional information

Used terms: 2, i