General rule for $(A \land B) \lor (\neg A \land D)$

Question

I encountered the following small expression: $$ (n\ge0\land y \gt 5) \lor(n \lt 0 \land x > 10). $$ The answer should be easily $(x > 10 \land y \gt 5)$ 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.

Answer 1

The proposition $(n\ge0\land y \gt 5) \lor(n \lt 0 \land x > 10)$ is not equivalent to $(x > 10 \lor y \gt 5)$.

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

However, it is true that $(n\ge0\land y \gt 5) \lor(n \lt 0 \land x > 10)$ implies $(x > 10 \lor y \gt 5)$.

Answer 2

Consider the cases $A = true$ and $\neg A = true$ and compare the truth values of these statements. So your answer should actually say $ x > 10 \vee y > 5$.

Answer 3

\begin{align*} (A \land B) \lor (\lnot A \land C) &\iff (A \lor \lnot A) \land (A \lor C) \land (B \lor \lnot A) \land (B \lor C) \\ &\iff (A \lor C) \land (\lnot A \lor B) \land (B \lor C) \end{align*} It's not hard to see that this implies $B \lor C$ but it is not equivalent to $B \lor C$.