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Proof step 15 from proof gcd-rec-sub-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: absa seq: ⊦ i ≤ 0 ∧ j ≤ 0 → abs(i + j) = abs(i) + abs(j) subst: [j, i] → [i, 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)% ⊦ ~ i + ~ j = ~(i + j); (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2% ⊦ ~ i - ~ j = j - i; (s)% ⊦ i - j + i = ~ j; (s)% ⊦ i - ~ j = i + j; (s)% ⊦ 0 < ~ i ↔ i < 0; (s)% ⊦ i < 0 ∧ 0 ≤ j → i < j; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < j → ¬ j < i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss) Used terms: i, i0