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Proof step 15 from proof gcd-com (previous step)

gcd(j, i) < gcd(i, j), gcd(i, j) | j ∧ gcd(i, j) | i ∧ (j ≠ 0 ∨ i ≠ 0) → gcd(i, j) ≤ gcd(j, i) ⊦

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 16:
gcd(j, i) < gcd(i, j) ⊦ gcd(i, j) | j ∧ gcd(i, j) | i ∧ (j ≠ 0 ∨ i ≠ 0)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (j ≤ i ↔ false); (as)