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Proof step 3 from proof gcd-divides-both-h (previous step)

i < j, j0 | gcd(i, j), 0 ≤ i, 0 < j, ∀ j1, i, j0. i→n(j1) < i→n(j) → 0 ≤ i ∧ i < j1 ∧ j0 | gcd(i, j1) → j0 | j1 ∧ j0 | i ⊦ j0 | j ∧ j0 | i

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 4:
⊦ gcd(i, j) = gcd(i, j - i)
Proof step 5:
i < j, j0 | gcd(i, j - i), 0 ≤ i, 0 < j, ∀ j1, i, j0. i→n(j1) < i→n(j) → 0 ≤ i ∧ i < j1 ∧ j0 | gcd(i, j1) → j0 | j1 ∧ j0 | i ⊦ j0 | j ∧ j0 | i

Rule argument

spec: int-gcd inst: name: gcd-rec-sub-01 seq: ⊦ gcd(i, j) = gcd(i, j - i) subst: [j, i] → [j, i] flags: rewrite, current

Additional information

Used terms: j, i