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Proof step 2 from proof div-mod-01 (previous step)

i0 ≤ i, 0 < j, 0 ≤ i0, 0 ≤ i ⊦ j | i - i0 ↔ i % j = i0 % j

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 3:
⊦ i | j ↔ (∃ i0. i * i0 = j)
Proof step 4:
i0 ≤ i, 0 < j, 0 ≤ i0, 0 ≤ i ⊦ (∃ i1. j * i1 = i - i0) ↔ i % j = i0 % j

Rule argument

spec: int-gcd inst: name: divides seq: ⊦ i | j ↔ (∃ i0. i * i0 = j) subst: [j, i] → [i - i0, j] flags: rewrite, current

Additional information

Used terms: i - i0, j