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I need to translate the following sentence:

Whenever the alarm sounds the fan must be on and the building empty

into a logical proposition.

So I defined variable as follow:

$p$ = "The alarm sounds"
$q$ = "The fan is on"
$r$ = "The building is empty"

The answer: $p \to (q ∧ r).$

I'm confused by this question because if the alarm does not sound and the fan is not on and the building is not empty then by implication doesn't this mean the statement is true?

amWhy
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vesta
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2 Answers2

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Note the three rows of truth value assignments for $p, q, r$ which yield "F" in the right most collomn.

We can summarize the matter as follows: $p\to (q\land r)$ is true, except when (p is T and ($q \land r$) is false), which happens when p is true, and either q is false, or r is false, or both q and r are false.

Simply put, we can simplify the statement as follows. Let $s:=q\land r$.

enter image description here

Then we have $p \to s$, which by definition is false if and only if $p$ is true, and $s$ is false.

Note that we are using the definition of the material conditional, in which nothing is needed in terms of relevancy, correlation, or cause-and effect relations between the $p,$ and $s$ in $p\to s$.

amWhy
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  • You might also want to check out the following links, too: https://math.stackexchange.com/questions/232309/how-to-interpret-material-conditional-and-explain-it-to-freshmen?noredirect=1&lq=1. – amWhy Jan 10 '18 at 16:32
  • More explanation can be found at this link. It discusses that since $a\to b$ is true, in all truth value assignments, except when ($a$ is true AND $b$ is false), in which case it is false. – amWhy Jan 10 '18 at 20:55
  • In other words we can recognize an implication $a\to b$ as true whenever $a$ is false OR whenever $b$ is true (one, or the other, or both holds). Whenever $a$ is false, no matter what $b$ is (we don't even need to know if it is true or false), the material implications is true. Likewise, when $b$ is true, the implication is true, regardless of whether $a$ is true or false. – amWhy Jan 10 '18 at 20:56
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Yup; in the case that the alarm does not sound, the fan is not on, and the building is not empty, the "Whenever..." sentence should come out true.

Indeed, $p \implies (q \wedge r)$ comes out true if $p$, $q$, and $r$ are all false.

BallBoy
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