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Proof step 59 from proof gcd-rec-sub-01 (previous step)

¬(0 < abs(i) ∧ abs(i) < abs(i0)), 0 < i, i0 < 0 ⊦ 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 60:
0 < i, i0 < 0 ⊦ 0 < abs(i) ∧ abs(i) < abs(i0), i0 = 0, gcd(abs(i), abs(i0 + i)) = gcd(abs(i), abs(i0))

Rule argument

pos: left 1

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