How do I prove that $$z_1 + x_{12} + x_{22} + x_{13} + x_{23} + z_4 \geq 2$$ is a valid inequality for my constraints: \begin{align} x_{11} + x_{12} + x_{13} &\geq 1 \tag1\label1\\ x_{12} + x_{13} + x_{14} &\geq 1 \tag2\label2\\ x_{21} + x_{22} + x_{23} + x_{24} &\geq 1 \tag3\label3\\ x_{ij} &\leq z_j &&\text{for all $i$ and $j$} \tag4\label4\\ x_{ij}, z_j &\in \{0,1\} &&\text{for all $i$ and $j$} \end{align}
I have tried several ways but the closest I have gotten is by taking \eqref{1} + \eqref{2} + 2\eqref{3} to get: $$x_{11} + 2x_{21} + 2x_{12} + 2x_{22} + 2x_{13} + 2x_{23} + x_{14} + 2x_{24} \geq 4$$ then replacing $x_{11}, x_{21}$ with $z_1$ and $x_{14}, x_{24}$ with $z_4$ and then dividing by 2 which gives the invalid inequality: $$\frac32 z_1 + x_{12} + x_{22} + x_{13} + x_{23} + \frac32 z_4 \geq 2$$
Adding up the first three constraints, $-x_{ij} + z_j \ge 0$ for $(i,j)\in\{1,2\}\times\{1,4\}$, and the lower bounds $x_{22} \ge 0$ and $x_{23} \ge 0$ yields $$2x_{12} + 2x_{13} + 2x_{22} + 2x_{23} + 2z_1 + 2z_4 \ge 3.$$ Now divide both sides by $2$ to obtain $$x_{12} + x_{13} + x_{22} + x_{23} + z_1 + z_4 \ge \frac{3}{2}.$$ Because the left-hand side is integer-valued, you can replace the right-hand side with $\lceil 3/2 \rceil = 2$, yielding the desired inequality.