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Proof step 1 from proof gcd-abs-rew-02 (conclusion)

⊦ gcd(i, j) = gcd(abs(i), abs(j))

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 2:
⊦ ¬ i < j → (i ≠ j ↔ j < i)
Proof step 3:
¬ ¬ gcd(i, j) < gcd(abs(i), abs(j)) ⊦ gcd(i, j) = gcd(abs(i), abs(j))
Proof step 18:
¬ gcd(i, j) < gcd(abs(i), abs(j)) ⊦ ¬ gcd(abs(i), abs(j)) < gcd(i, j)

Rule argument

spec: int-basic1 inst: name: lst-01 seq: ⊦ ¬ i < j → (i ≠ j ↔ j < i) subst: [j, i] → [gcd(abs(i), abs(j)), gcd(i, j)] flags: rewrite

Additional information

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