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Proof step 45 from proof gcd-rec-sub-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 46:
⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(j, j - i)

Rule argument

spec: <toplevel> name: gcd-rec-sub-02 seq: ⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(j, j - i) subst: [j, i] → [abs(i), abs(i0)] flags: <all paths>

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ 0 < j → abs(j) = j; (s)% ⊦ ~ i - j = ~(i + j); (s)% ⊦ i < ~ j ↔ i + j < 0; (s) Used terms: abs(i), abs(i0)