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Proof step 65 from proof gcd-rec-sub-01 (previous step)

abs(i) = abs(i0), 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

Rule argument

fma: abs(i0 + i) = 0

Additional information

Used simplifier rules: abs-16: ⊦ 0 < j → abs(j) = j; from specification int-abs%abs-08: ⊦ i < 0 → abs(i) = ~ i; from specification int-abs%negeq: ⊦ ~ i = j ↔ i + j = 0; from specification int-abs%abs-07: ⊦ abs(i) = 0 ↔ i = 0; from specification int-abs Used formulas: abs(i0 + i) = 0