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

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

• Shouldn't it be $(x > 10 \lor y > 5)?$ Mar 10, 2019 at 14:56
• Mh it may be. Would you be so kind to tell me which rules did you apply in order to get there? Mar 10, 2019 at 15:05
• I just say, one of the statements $n\ge0$ and $n<0$ must be true so one of the two clauses must be true. In the first case,$x>10.$ In the second, $y>5.$ I don't think there's much use for symbolic logic in ordinary reasoning. However, you can apply the distributives laws a couple of times. Mar 10, 2019 at 15:08

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 $$(x > 10 \lor y \gt 5)$$.
\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 $$B \lor C$$.
Consider the cases $$A = true$$ and $$\neg A = true$$ and compare the truth values of these statements. So your answer should actually say $$x > 10 \vee y > 5$$.
• What if $A$ and $C$ are true, while $B$ is false ? Then the expression $(A \wedge B) \vee (\lnot A \wedge C)$ is false but $B \vee C$ is true. Mar 10, 2019 at 23:36