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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.

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    $\begingroup$ Shouldn't it be $(x > 10 \lor y > 5)?$ $\endgroup$
    – saulspatz
    Mar 10, 2019 at 14:56
  • $\begingroup$ Mh it may be. Would you be so kind to tell me which rules did you apply in order to get there? $\endgroup$ Mar 10, 2019 at 15:05
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    $\begingroup$ 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. $\endgroup$
    – saulspatz
    Mar 10, 2019 at 15:08

3 Answers 3

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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 $(x > 10 \lor y \gt 5)$.

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\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 $B \lor C$.

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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$.

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    $\begingroup$ 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. $\endgroup$
    – tristan
    Mar 10, 2019 at 23:36
  • $\begingroup$ Sorry, you are right. I should have said implication instead. I have edited my comment. $\endgroup$
    – vxnture
    Mar 11, 2019 at 9:16

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