Why $((+ 1) ((+ 1) 1))$ is not a $\delta-$redex?

Question

I was reading http://barrywatson.se/lsi/lsi_delta_reduction.html, where

((+ 1) 2) →δ 3 is a δ-reduction.
((+ 1) ((+ 1) 1)) is a not a δ-redex.

Wouldn't $((+ 1) ((+ 1) 1)) \rightarrow ((+1) 2)) \rightarrow 3$?

Answer 1

Because the definition of $\delta$-reduction requires that

for all $1≤i≤n$, $N_i$ is normalized

and the second argument $((+1) 1)$ is not normalized.

As you observe, we can first reduce the $\delta$-redex $((+1) 1)$ to $\delta$-normal form $2$; then $((+1) 2)$ is a $\delta$-redex which can be reduced to $\delta$-normal form $3$.
$((+ 1) ((+ 1) 1)) \triangleright_\delta ((+1) 2)) \triangleright_\delta 3$ is a correct reduction, but $((+ 1) ((+ 1) 1))$ isn't a redex in the first step; $((+ 1) 1)$ is.