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Proof step 19 from proof gcd-rec-sub-01 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Rule argument

spec: int-abs name: Abs-neg seq: ⊦ j < 0 → abs(j) = 0 - j subst: [j] → [i0 + i] flags: <all paths>

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);% ⊦ 0 - i = ~ i; (s)% ⊦ i + ~ j = i - j; (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% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ i + i = 2 * i; (s)% ⊦ i < 0 ∧ j < 0 → i + j < 0; (s) Used terms: i0 + i