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Proof step 31 from proof gcd-rec-sub-01 (previous step)

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

User applied rule case distinction

And obtained the following new goals

Proof step 32:
abs(i) = abs(i0), 0 < i0, i < 0 ⊦ gcd(i, i0 - abs(i)) = gcd(abs(i), abs(i0)), i = 0
Proof step 35:
0 < i0, i < 0, abs(i0) < abs(i) ⊦ gcd(i, i0 - abs(i)) = gcd(abs(i), abs(i0)), i = 0

Rule argument

pos: left 3

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