Identifying redexes in a lambda expression I've been given the following expression and asked to identify all redexes
\begin{align}
(x.xx)(x.xx)(x.xx)(x.xx)
\end{align}
The example key gives the first part of the expression $(x.xx)(x.xx)$.
I don't understand why the second part of the expression is not also a redex as it follows the form $(λx.t)t’$. 
Would the answer not be $(x.xx)(x.xx)$ and $(x.xx)(x.xx)$ as the expression breaks down into 
$(λx.t)t’(λx.t)t’$? 
 A: In the $\lambda$-calculus, a $\lambda$-term of the form $t_1t_2t_3t_4$ must be read as 
\begin{align}
 ((t_1 t_2)t_3)t_4
\end{align}
i.e. the application $tu$ of the $\lambda$-term $t$ to the $\lambda$-term $u$ is left-associative.
According to this convention, in your $\lambda$-term $(\lambda x_1 \, x_1x_1) (\lambda x_2 \, x_2x_2) (\lambda x_3 \, x_3x_3) (\lambda x_4 \, x_4x_4)$ there is only one redex, $(\lambda x_1 \, x_1x_1) (\lambda x_2 \, x_2x_2)$, as correctly stated in the example you mentioned. 
Indeed, your $\lambda$-term is (pay attention to the parentheses)
\begin{align}
\Big(\big((\lambda x_1 \, x_1x_1) (\lambda x_2 \, x_2x_2)\big) (\lambda x_3 \, x_3x_3)\Big) (\lambda x_4 \, x_4x_4)
\end{align}
therefore, there is no redex $(\lambda x_3 \, x_3x_3) (\lambda x_4 \, x_4x_4)$ in your $\lambda$-term because $(\lambda x_3 \, x_3x_3) (\lambda x_4 \, x_4x_4)$ is not a sub-term of your $\lambda$-term: in fact, $\lambda x_3 \, x_3x_3$ is actually the argument of another application. 
Similarly for $(\lambda x_2 \, x_2x_2) (\lambda x_3 \, x_3x_3)$.
