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Proof step 6 from proof gcd-com (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
gcd(i, j) < gcd(j, i) ⊦ gcd(i, j) = gcd(j, i), gcd(j, i) | i ∧ gcd(j, i) | j

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → ¬ j < i; (s)% ⊦ i ≠ j → (i ≤ j ↔ i < j); (s)