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Proof step 34 from proof gcd-rec-sub-01 (previous step)

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

User applied rule insert equation

And obtained the following new goals

Proof step 35:
abs(i) = abs(i0), i < 0, 0 < i0 ⊦ gcd(i, i0 - abs(i0)) = gcd(abs(i0), abs(i0)), i = 0

Rule argument

pos: left 1 flags:

Additional information