Let $X$, $Y$, $Z$, $A$ be a set of random variables drawn from the Irwin-Hall distribution where $X$ is the sum of $c$ iid r.v.s, $Y$ is the sum of $c$ iid r.v., $Z$ is the sum of $n - c$ iid r.v.s, and $A$ is the sum of $n - c$ iid r.v.s. 


I want to compare $\Pr[(X \leq x) \cap (Z \leq x - X) \cap (A \leq x - X)]$ with $\Pr[(X \leq x) \cap (Y \leq x) \cap (Z \leq x - X) \cap (A \leq x - Y)]$. Intuitively, it seems like $\Pr[(X \leq x) \cap (Z \leq x - X) \cap (A \leq x - X)] \geq \Pr[(X \leq x) \cap (Y \leq x) \cap (Z \leq x - X) \cap (A \leq x - Y)]$ but I couldn't find a clean proof for this.


You can assume that $x \in [0, n]$. 



Irwin-Hall Distribution relationship between two sets of events
$\newcommand{\R}{\mathbb{R}}$
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Your conditions on $X$, $Y$, $Z$, $A$, as I understood them, imply that $X$, $Y$, $Z$, $A$ are independent nonnegative random variables (r.v.'s), with $(X,Z)$ equal $(Y,A)$ in distribution -- which is all we need to verify your conjecture. 


Indeed, since $Z,A\ge0$, we see that event $\{Z \le x - X\}$ implies $\{X\le x\}$, and $\{A \le x - Y\}$ implies $\{Y\le x\}$. So, denoting by $F$ the common cdf of the iid r.v.'s $Z$ and $A$, we see that your conjecture simplifies to $p\_1\ge p\_2$, where 
\begin{equation}
 p\_1:=P(X+Z\le x,X+A\le x)=\int\_\R P(X\in du)P(Z\le x-u,A\le x-u)
 =\int\_\R P(X\in du)F(x-u)^2=EF(x-X)^2, 
\end{equation}
\begin{equation}
 p\_2:=P(X+Z\le x,Y+A\le x)=P(X+Z\le x)^2=
 \Big(\int\_\R P(X\in du)P(Z\le x-u)\Big)^2
=\big(EF(x-X)\big)^2. 
\end{equation}
So, by the Cauchy--Schwarz inequality, we indeed have $p\_1\ge p\_2$. 


