# $a,b,c,d$ are independent random variables, then $a+b$, $c+d$ are independent

By looking at the characteristic function of $(a,b,c,d)$, from independence we have $$\mathbb{E} e^{i(ua+vb+wc+zd)} = \mathbb{E}e^{iua} \mathbb{E}e^{ivb} \mathbb{E}e^{iwc}\mathbb{E}e^{izd} = \mathbb{E}e^{i(ua+vb)} \mathbb{E}e^{i(wc+zd)}$$ this implies $$\mathbb{E} e^{i(u(a+b)+v(c+d))} = \mathbb{E}e^{iu(a+b)} \mathbb{E}e^{iv(c+d)}$$ which implies $a+b$ and $c+d$ are independent.

Is this correct, and is there a more direct argument?

a) Yes, this is correct by Kac's theorem; see here: Moment generating functions/ Characteristic functions of $X,Y$ factor implies $X,Y$ independent.
b) You can look at theorem 2.1.6 from Durrett's book on Probability Theory and Examples (4th Edition) (and cite it; it becomes very direct then!); it depends on what definition of independence you want to use. There is a bunch of answers on StackExchange on various other proofs; here is one $X$,$Y$,$Z$ mutually independent implies $X+Y$ independent of $Z$