Is the similar argument presented in the proof below correct: " if $\ln \Gamma(z) + \ln \Gamma(1-z)$ is algebraic for some rational not integer $z$ , then, so it's their difference, $\ln \Gamma(z) - \ln \Gamma(1-z)$ " ?
In the proof below they use distinct values for $x$, but let's keep just one variable $z$.
The question here is: if $a,b$ are real numbers and if we assume that the sum $a+b$ is algebraic, then so it is $a-b$. If both $a,b \in \overline{\mathbb{Q}}$ then this is true. And there is exceptions like the example commented below.
This is not necessarily true. Take $a=\pi$ and $b=-\pi$ for instance.
Edit
In response to the proof you added to the question, here is your error: They have not claimed that if $a+b$ is algebraic then $a-b$ is algebraic. They have used the fact that if $a$ and $b$ are both algebraic, then $a-b$ is algebraic, which is true.