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Proof step 48 from proof gcd-rec-sub-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 49:
⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(i, j - i)

Rule argument

spec: <toplevel> name: gcd-rec-sub seq: ⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(i, 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);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ~ i - j = ~(i + j); (s)% ⊦ ~ j < ~ i ↔ i < j; (s)% ⊦ i < i + j ↔ 0 < j; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ 0 < j → abs(j) = j; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < ~ j ↔ i + j < 0; (s) Used terms: abs(i), abs(i0)