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Proof step 81 from proof gcd-rec-sub-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 82:
abs(i0 + i) = abs(i0) - abs(i), i0 < i, 0 < i, i0 < 0 ⊦ abs(i0) < abs(i), i0 = 0, i = 0, gcd(abs(i), abs(i0 + i)) = gcd(abs(i0 + i), abs(i0))

Rule argument

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

Additional information

Used simplifier rules: gcd-neg: ⊦ gcd(~ i, j) = gcd(i, j);%gcd-com: ⊦ gcd(i, j) = gcd(j, i);%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%subsub: ⊦ ~ i - j = ~(i + j); from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abseq: ⊦ abs(i) + i = 0 ↔ i ≤ 0; from specification int-abs%lels-01: ⊦ ¬ i ≤ j ↔ j < i; from specification int-abs%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2%neg-13: ⊦ ~ i < j ↔ 0 < j + i; from specification int-abs%lels: ⊦ ¬ i < j ↔ j ≤ i; from specification int-abs%lels-03: ⊦ i < j → (¬ ¬ j ≤ i ↔ false); from specification int-abs Used formulas: abs(i0 + i) = abs(i0) - abs(i)