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Proof step 66 from proof gcd-rec-sub-01 (
previous step
)
abs(i0 + i) = 0, abs(i) = abs(i0), 0 < i, i0 < 0 ⊦ i0 = 0, gcd(abs(i), abs(i0 + i)) = gcd(abs(i), abs(i0))
User applied rule insert equation
And obtained the following new goals
Proof step 67
:
abs(i) = abs(i0), 0 < i, i0 < 0 ⊦ i0 = 0, gcd(abs(i), 0) = gcd(abs(i), abs(i0))
Rule argument
pos: left 1 flags: discard
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