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

gcd(i, j) < gcd(j, i) ⊦ gcd(i, j) = gcd(j, i)

User applied rule insert lemma

And obtained the following new goals

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

Rule argument

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

Additional information

Used terms: gcd(j, i), i, j