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Proof step 21 from proof gcd-abs-rew-02 (previous step)

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

Heuristic simplifier applied rule simplifier

And obtained the following new goals

Proof step 22:
gcd(abs(i), abs(j)) < gcd(i, j) ⊦ gcd(i, j) | abs(i) ∧ gcd(i, j) | abs(j) ∧ (i ≠ 0 ∨ j ≠ 0)

Rule argument

Additional information

Used simplifier rules: ⊦ i < j → (j ≤ i ↔ false); (as)% ⊦ abs(i) = 0 ↔ i = 0; (s)