General rule for $(A \land B) \lor (\neg A \land D)$ I encountered the following small expression: 
$$
(n\ge0\land y \gt 5) \lor(n \lt 0  \land x &gt; 10).  
$$
The answer should be easily $(x &gt; 10 \land y \gt 5)$ but unfortunately I don't see how the components get factored out in a way to exploit the tautology $n\ge0 \lor n\lt 0 = true$ and simplify the expression.
I tried to use the distributive law but I got a longer and worse expression. 
Maybe there is a general proven formula to back up the computation in these cases?

Any help would be extremely appreciated.
The proposition $(n\ge0\land y \gt 5) \lor(n \lt 0  \land x &gt; 10)$ is not equivalent to $(x &gt; 10 \lor y \gt 5)$.

For example, take $n=0$, $x=11$ and $y=0$.

However, it is true that $(n\ge0\land y \gt 5) \lor(n \lt 0  \land x &gt; 10)$  implies $(x &gt; 10 \lor y \gt 5)$.
Consider the cases $A = true$ and $\neg A = true$ and compare the truth values of these statements. So your answer should actually say $ x &gt; 10 \vee y &gt; 5$.