= Solution
Label the measured <vertex> of the four-cycle by $0$, its two neighbours by $1,3$, and the opposite <vertex> by $2$. The <edges> are $01,12,23,30$. By the definition of a <graph state>,
$$
|\psi_{2\times2}\rangle=E_{01}E_{12}E_{23}E_{30}|+\rangle^{\otimes4}.
$$
All <Controlled-Z gates> commute. Conditioning on the <computational basis> value $r$ of <vertex> $0$, the factors on <edges> $01$ and $03$ act on their other endpoints as $Z_1^r$ and $Z_3^r$, since $E|r,a\rangle=(-1)^{ra}|r,a\rangle$. The remaining <edges> form the three-<vertex> path $1-2-3$:
$$
{}_0\langle r|\psi_{2\times2}\rangle=\frac1{\sqrt2}Z_1^rZ_3^rE_{12}E_{23}|+\rangle_1|+\rangle_2|+\rangle_3.
$$
The <Born rule> gives probability one half for either result. Normalizing proves
$$
\boxed{|\psi_{\mathrm{remaining}}\rangle=(Z^r\otimes I\otimes Z^r)|\psi_3\rangle.}
$$
This is <computational-basis measurement of a graph-state vertex>: remove the measured <vertex> and apply a known <Pauli Z gate> byproduct to each neighbour. The chosen labels specify precisely which factors occur; any <vertex> of the square can serve as <vertex> $0$ by relabelling.
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