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Proof step 28 from proof gcd-abs-rew-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Rule argument

Additional information