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Proof step 6 from proof gcd-rec (previous step)

abs(i) = i0 * abs(j) + j0, 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ j = 0, gcd(abs(i), abs(j)) = gcd(abs(j), j0)

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 7:
⊦ ¬ i < j → (i ≠ j ↔ j < i)
Proof step 8:
¬ ¬ gcd(abs(i), abs(j)) < gcd(abs(j), j0), abs(i) = i0 * abs(j) + j0, 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ j = 0, gcd(abs(i), abs(j)) = gcd(abs(j), j0)
Proof step 35:
¬ gcd(abs(i), abs(j)) < gcd(abs(j), j0), abs(i) = i0 * abs(j) + j0, 0 ≤ i0, 0 ≤ j0, j0 < abs(j) ⊦ j = 0, ¬ gcd(abs(j), j0) < gcd(abs(i), abs(j))

Rule argument

spec: int-basic1 inst: name: lst-01 seq: ⊦ ¬ i < j → (i ≠ j ↔ j < i) subst: [j, i] → [gcd(abs(j), j0), gcd(abs(i), abs(j))] flags: rewrite

Additional information

Used terms: gcd(abs(j), j0), gcd(abs(i), abs(j))