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Proof step 9 from proof gcd-abs-rew-02 (previous step)

¬ gcd(abs(i), abs(j)) | abs(i), 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)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 10:
gcd(i, j) < gcd(abs(i), abs(j)) ⊦ gcd(i, j) = gcd(abs(i), abs(j)), gcd(abs(i), abs(j)) | abs(i), gcd(abs(i), abs(j)) | i ∧ gcd(abs(i), abs(j)) | j ∧ (i ≠ 0 ∨ j ≠ 0)

Proof step 9 from proof gcd-abs-rew-02 (previous step)

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

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