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Proof step 10 from proof gcd-rec-sub-03 (previous step)

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

User applied rule cut formula

And obtained the following new goals

Proof step 11:
i0 < 0, i < 0 ⊦ gcd(abs(i), abs(i0 + i)) = gcd(i0 + i, i0), 0 < i ∧ i < i0 + i
Proof step 17:
i < 0 ⊦ i0 < 0, gcd(abs(i), abs(i0 + i)) = gcd(i0 + i, i0), 0 < i ∧ i < i0 + i

Rule argument

fma: i0 < 0

Additional information

Used formulas: i0 < 0