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The question I am working on is:

Let $p$ and$q$ be the propositions

$p$:It is below freezing.

$q$:It is snowing. Write these propositions using $p$ and $q$ and logical connectives (including negations).

a) It is below freezing and snowing.

b)It is below freezing but not snowing.

c) It is not below freezing and it is not snowing.

d)It is either snowing or below freezing (or both).

e) If it is below freezing, it is also snowing.

f) Either it is below freezing or it is snowing, but it is not snowing if it is below freezing.

g) That it is below freezing is necessary and sufficient for it to be snowing."

I only actually have a problem with part d). The answer I wrote is $(p \vee q)\vee (p \wedge q)$; but, apparently, the answer is simply $(p \vee q)$. What I'd like to know if my answer is indeed correct, but just redundant.

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It's just redundant: describes the same set in more words. :) –  gnometorule Dec 11 '12 at 14:53
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Well maybe to explain a bit more: A or B means (A but not B) or (B but not A) or (A and B) (just as in plain English); so the right parenthesized term is already in the left par. term. Adding it with 'or' will not harm, but also not change the set. –  gnometorule Dec 11 '12 at 14:59
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1 Answer

up vote 3 down vote accepted

As gnometorule said, your extra term is simply redundant.

Informally, note that if $p\land q$ is true, then both $p$ and $q$ are true, so $p\lor q$ is certainly true; thus, allowing the extra possibility $p\land q$ in addition to $p\lor q$ adds nothing. More formally,

$$\begin{align*} (p\lor q)\lor(p\land q)&\equiv p\lor\Big(q\lor(p\land q)\Big)&&\text{associativity}\\ &\equiv p\lor\Big(q\lor(q\land p)\Big)&&\text{commutativity}\\ &\equiv p\lor q&&\text{absorption}\;. \end{align*}$$

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