CSS encoding from nested binary codes

The displayed generator rows of $C_1$ are independent, and the single generator of $C_2$ is nonzero and lies in $C_1$. Thus $\dim C_1=2$, $\dim C_2=1$, $k=1$, and the two displayed cosets are $C_2$ and $1100+C_2$. Their supports are disjoint, so [F1] gives the orthonormal logical basis shown.

A vector $y=(y_1,y_2,y_3,y_4)$ lies in $C_1^{\perp }$ exactly when $y_1+y_2+y_3+y_4=0$ and $y_1+y_2=0$. These equations are equivalent to $y_1=y_2$ and $y_3=y_4$, so $C_1^{\perp }=\mathrm{span}\{1100,0011\}$. Both generators have even dot product with $1111$, so the X and Z checks commute. The vector $1010$ has even weight and hence lies in $C_2^{\perp }$, but it is not one of $0000,1111,1100,0011$, so it is not in $C_1$. This verifies $C_2^{\perp }\subseteq ̸C_1$.

Translation by $1100$ exchanges the two displayed logical coset states, so $X^{1100}$ acts as logical bit flip. The phase $(-1{)}^{1010\cdot y}$ is $+1$ for $y=0000,1111$ and $-1$ for $y=1100,0011$, so $Z^{1010}$ acts as logical phase flip. Their dot product is $1100\cdot 1010=1$ in ${\mathbb{F}}_2$, hence the logical operators anticommute.