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Proof step 23 from proof gcd-rec-sub-01 (previous step)

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

User applied rule insert equation

And obtained the following new goals

Proof step 24:
i < 0, 0 < i0 ⊦ gcd(i, i0 - abs(i)) = gcd(i, i0), i = 0

Rule argument

pos: left 1 flags: discard

Additional information