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Proof step 7 from proof gcd-rec (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 8:
⊦ ¬ i < j → (i = j ↔ ¬ j < i)
Proof step 9:
abs(i) = i0 * abs(j) + j0, gcd(abs(i), abs(j)) < gcd(abs(j), j0), j0 < abs(j), 0 ≤ i0, 0 ≤ j0 ⊦ gcd(abs(i), abs(j)) = gcd(abs(j), j0), j = 0
Proof step 36:
abs(i) = i0 * abs(j) + j0, gcd(abs(j), j0) < gcd(abs(i), abs(j)), j0 < abs(j), 0 ≤ i0, 0 ≤ j0 ⊦ j = 0, gcd(abs(i), abs(j)) < gcd(abs(j), j0)

Rule argument

spec: int-basic1 name: lst-01 seq: ⊦ ¬ i < j → (i = j ↔ ¬ j < i) subst: [j, i] → [gcd(abs(j), j0), gcd(abs(i), abs(j))] flags: <all paths>

Additional information

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