# Are standalone propositions affected by negation operators?

In propositional logic, for example: $$\neg p \vee q.$$

If $p$ is true at the outset, does that mean it must be considered false when comparing with q in the disjunction?

P.S. I am unsure about tags for this question.

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@Arturo: Not criticizing, just curious: Why didn't you find the logic tag appropriate? – Mike Spivey Nov 29 '10 at 3:20
@Mike Spivey: Oops; I just typed in the one that seemed obvious, did not realize I had obliterated an adequate one as well. I think I accidentally "highlighted-deleted" it when I selected the tag bar to put in (propositional-calculus) – Arturo Magidin Nov 29 '10 at 3:23

If $p$ is true, then $\neg p$ is false. To evaluate $\neg p \vee q$, you must evaluate $\neg p$ and you must evaluate $q$. If either $\neg p$ is true or $q$ is true, then $\neg p\vee q$ is true.

In other words, you really need to figure out $(\neg p)\vee q$, performing first the operation inside the parentheses, then the disjunction.

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If $p$ is true, then $\lnot p \lor q \Leftrightarrow q$.

In general, $p$ and $\lnot p$ have the opposite value: if one is true then the other is false, and vice versa.

You can think of $p$ as some proposition, say "today is Sunday". Then $\lnot p$ stands for "today is not Sunday".

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I'm not sure I understand your question, but this may help.

Truth Table for ~p v q:

~ p v q
F T T T
F T F F
T F T T
T F T F

If p is true, and ~p v q is true (first line only), then q is true.

Note that ~p v q is logically equivalent to p => q.

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