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Proof step 72 from proof gcd-rec-sub-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 73:
abs(i0 + i) = abs(i) - abs(i0), abs(i0) < abs(i), 0 < i, i0 < 0 ⊦ i0 = 0, gcd(abs(i), abs(i0 + i)) = gcd(abs(i), abs(i0))

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);%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%subneg: ⊦ i - ~ j = i + j; from specification int-abs%add-c: ⊦ i + j = j + i; from specification int-basic2%abs-18: ⊦ abs(i) = i ↔ 0 ≤ i; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%ls-01: ⊦ i < j → ¬ j < i; from specification int-basic1 Used formulas: abs(i0 + i) = abs(i) - abs(i0)