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Proof step 24 from proof gcd-rec-sub-03 (previous step)

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

User applied rule insert rewrite lemma

And obtained the following new goals

Proof step 25:
⊦ 0 < i ∧ i < j → gcd(i, j) = gcd(j, j - i)
Proof step 26:
i < 0, abs(i) < i0, 0 < i0 - abs(i), i0 - abs(i) < abs(i0) ⊦ gcd(abs(i), i0 - abs(i)) = gcd(abs(i0), abs(i0) - (i0 - abs(i))), i0 < 0, 0 < i ∧ i < i0 + 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(i0), i0 - abs(i)] flags: <all paths>

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ i - ~ j = i + j; (s)% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ 0 ≤ i → abs(i) = i; (s)% ⊦ i + j < j ↔ i < 0; (s)%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used terms: abs(i0), i0 - abs(i)