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Proof step 68 from proof gcd-rec-sub-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 69:
abs(i0 + i) = abs(i) - abs(i0), abs(i0) < abs(i), 0 < i, i0 < i, i0 < 0 ⊦ gcd(abs(i), abs(i0 + i)) = gcd(abs(i0 + i), abs(i0)), i = 0, i0 = 0

Rule argument

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

Additional information

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