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Proof step 34 from proof gcd-rec-sub-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 35:
abs(i0 + i) = abs(i) - abs(i0), i < 0 ⊦ gcd(abs(i), abs(i0 + i)) = gcd(abs(i0 + i), abs(i0)), i0 < 0, abs(i) < i0, 0 < i ∧ i < i0 + i

Rule argument

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

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);% ⊦ i < 0 → abs(i) = ~ i; (s)% ⊦ ¬ j < 0 → abs(j) = j; (s)% ⊦ ~ i - j = ~(i + j); (s)% ⊦ ~ i = j ↔ i + j = 0; (s)% ⊦ abs(i) + i = 0 ↔ i ≤ 0; (s)% ⊦ i ≤ j ↔ ¬ j < i; (ss)%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)% ⊦ ~ i < j ↔ 0 < j + i; (s)% ⊦ i < j ↔ ¬ j ≤ i; (ss)% ⊦ i < j → (j ≤ i ↔ false); (as) Used formulas: abs(i0 + i) = abs(i) - abs(i0)