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Proof step 1 from proof div-mod (conclusion)

⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j | i - i0 ↔ i % j = i0 % j)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i → (j | i - i0 ↔ i % j = i0 % j)
Proof step 3:
¬(0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i) ⊦ 0 < j ∧ 0 ≤ i ∧ 0 ≤ i0 → (j | i - i0 ↔ i % j = i0 % j)

Rule argument

spec: int-gcd inst: name: div-mod-01 seq: ⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i → (j | i - i0 ↔ i % j = i0 % j) subst: [i0, i, j] → [i0, i, j] flags: rewrite, current

Additional information

Used terms: i0, i, j