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Proof step 22 from proof gcd-rec-sub-01 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 23:
i0 + i = i0 - abs(i), i < 0, 0 < i0 ⊦ gcd(i, i0 + i) = gcd(i, i0), i = 0

Rule argument

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

Additional information

Used simplifier rules: abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%subneg: ⊦ i - ~ j = i + j; from specification int-abs%add-in: ⊦ i + j = i + i0 ↔ j = i0; from specification int-basic2%add-a: ⊦ i + j + i0 = (i + j) + i0; from specification int-basic2%add-c: ⊦ i + j = j + i; from specification int-basic2 Used formulas: i0 + i = i0 - abs(i)