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Proof step 63 from proof gcd-rec-sub-01 (previous step)

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

User applied rule prop simplification

And obtained the following new goals

Proof step 64:
abs(i) = abs(i0) ∨ abs(i0) < abs(i), 0 < i, i0 < 0 ⊦ i0 = 0, gcd(abs(i), abs(i0 + i)) = gcd(abs(i), abs(i0))

Rule argument

pos: right 1

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