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Proof step 19 from proof gcd-abs-rew-02 (previous step)

gcd(abs(i), abs(j)) < gcd(i, j) ⊦

User applied rule insert lemma

And obtained the following new goals

Proof step 20:
⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j)
Proof step 21:
gcd(i, j) | abs(i) ∧ gcd(i, j) | abs(j) ∧ (abs(i) ≠ 0 ∨ abs(j) ≠ 0) → gcd(i, j) ≤ gcd(abs(i), abs(j)), gcd(abs(i), abs(j)) < gcd(i, j) ⊦

Rule argument

spec: int-gcd inst: name: gcd-max seq: ⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j) subst: [i0, i, j] → [gcd(i, j), abs(i), abs(j)] flags: current, []

Additional information

Used terms: gcd(i, j), abs(i), abs(j)