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Proof step 2 from proof div-mod (previous step)

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 3:
⊦ 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i → (j | i - i0 ↔ i % j = i0 % j)
Proof step 4:
0 < j, 0 ≤ i0, 0 ≤ i ⊦ j | i - i0 ↔ i % j = i0 % j, 0 < j ∧ 0 ≤ i0 ∧ i0 ≤ i

Rule argument

spec: <toplevel> 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: <all paths>

Additional information

Used terms: i0, i, j