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Proof step 26 from proof gcd-rec-sub-01 (previous step)

i < 0, 0 < i0 ⊦ gcd(i, i0 - abs(i)) = gcd(abs(i), abs(i0)), i = 0

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 27:
⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(i, j - i)
Proof step 28:
¬(0 < abs(i) ∧ abs(i) < abs(i0)), i < 0, 0 < i0 ⊦ gcd(i, i0 - abs(i)) = gcd(abs(i), abs(i0)), i = 0

Rule argument

spec: int-gcd inst: name: gcd-rec-sub seq: ⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(i, j - i) subst: [j, i] → [abs(i0), abs(i)] flags: rewrite, current

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%subneg: ⊦ i - ~ j = i + j; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs Used terms: abs(i0), abs(i)