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Proof step 18 from proof gcd-com (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 19:
gcd(j, i) < gcd(i, j) ⊦ gcd(i, j) | i ∧ (j ≠ 0 ∨ i ≠ 0)

Rule argument

Additional information