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Proof step 18 from proof gcd-rec-sub-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 19:
abs(i0 + i) = i0 - abs(i), abs(i) < i0, i < 0 ⊦ gcd(abs(i), abs(i0 + i)) = gcd(i0 + i, i0), i0 < 0, 0 < i ∧ i < i0 + i

Rule argument

fma: abs(i0 + i) = i0 - abs(i)

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)% ⊦ abs(i) = i ↔ 0 ≤ i; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < j → ¬ j < i; (s) Used formulas: abs(i0 + i) = i0 - abs(i)