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Proof step 4 from proof gcd-abs-rew-02 (previous step)

gcd(i, j) < gcd(abs(i), abs(j)) ⊦ gcd(i, j) = gcd(abs(i), abs(j))

User applied rule insert lemma

And obtained the following new goals

Proof step 5:
⊦ i0 | i ∧ i0 | j ∧ (i ≠ 0 ∨ j ≠ 0) → i0 ≤ gcd(i, j)
Proof step 6:
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)) ⊦ gcd(i, j) = gcd(abs(i), abs(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(abs(i), abs(j)), i, j] flags: current, []

Additional information

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