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

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

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

spec: int-gcd inst: name: gcd-less seq: ⊦ i1 | i ∧ i1 | j ∧ 0 ≠ i → i1 ≤ gcd(i, j) subst: [i1, i, j] → [i0, i, j] flags: current, []

Additional information

Used terms: i0, i, j