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Proof step 4 from proof gcd-one (previous step)

gcd(i, j) = 1, 0 < i0, 1 < i0, i0 | i, i0 | j ⊦

User applied rule apply lemma

And obtained the following new goals

Proof step 5:
⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j)
Proof step 6:
gcd(i, j) = 1, 1 < i0, 0 < i0, i0 | j, i0 | i, i0 | i ∧ i0 | j ∧ 0 ≠ i → i0 ≤ gcd(i, j) ⊦

Rule argument

spec: <toplevel> name: gcd-less seq: ⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j) subst: [i1, i, j] → [i0, i, j]

Additional information

Used terms: i0, i, j