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Proof step 6 from proof gcd-abs-rew-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 7:
gcd(i, j) < gcd(abs(i), abs(j)) ⊦ gcd(i, j) = gcd(abs(i), abs(j)), gcd(abs(i), abs(j)) | i ∧ gcd(abs(i), abs(j)) | j ∧ (i ≠ 0 ∨ j ≠ 0)

Rule argument

Additional information

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