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I'm struggling understanding truth tables.

Let's denote a true proposition by 1 and a false proposition by 0. We will be considering the propositional operation, $\Rightarrow$ (implies).

The truth table looks like the following

\begin{array}{c|cc}\rightarrow & 0 & 1\\\hline\\0 & 1 & 1\\1 & 0 & 1\end{array}

Does this say that $(0 \land 0) \rightarrow 1$? If it does, what exactly does that mean -- are we looking at $(0\rightarrow0)$ as one proposition and saying it is true?

Can you please explain this in english -- using sentences like (x>5 for propositions instead of just writing $P$ or $Q$ for propositions)

EDIT

amWhy
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Adeeb
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3 Answers3

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What you've posted is the truth table for material implication (the conditional) $p \rightarrow q$.

EDIT:

To better understand the material conditional (the connective $\rightarrow$, i.e. implication), see the following posts:

At each of those links, you'll find more linked questions that are also relevant. You are not alone: logical implication (e.g. $p\rightarrow q$) is perhaps the most difficult connective to grasp, in terms of its truth-table and how it is defined, in classical logic, (which is, in part, explained by the fact that in natural language, the term "implies" is used in ways whose meaning is not captured by its narrower meaning, as defined in logic).

If you have any further questions, I'll be happy to try and answer them!

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amWhy
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  • I was raised under the terminology that material implication is a connective between formulas (or propositions) and logical implication is a statement about formulas (or propositions). – Asaf Karagila Feb 12 '13 at 21:49
  • @Asaf: Yes, I customarily do the same, but it seems the convention isn't universally followed (or, perhaps the distinction is not universally understood). – amWhy Feb 12 '13 at 21:53
  • Convincing answer Amy. :+) – Mikasa Mar 03 '13 at 12:29
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See this answer, or any of the other fine answers to that question.

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Cameron Buie
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  • Cameron - the op clarified that he meant $\rightarrow$: implication. – amWhy Feb 12 '13 at 21:03
  • @amWhy: Thanks for letting me know. – Cameron Buie Feb 12 '13 at 21:08
  • In any standard logical language, $\land$ is a propositional connective, and can only connect propositional clauses. In such languages, $(P_1\land Q_1)(x)$ is therefore ill-formed. – Peter Smith Feb 12 '13 at 21:10
  • @Peter Smith, can you please simplify that sentence. I don't understand what you mean. – Adeeb Feb 12 '13 at 21:14
  • @PeterSmith: Fair enough. Here, though, $P_1,Q_1$ are functions $\Bbb Z\to{0,1}$ defined by the truth or falsity of the corresponding statement, and $\land$ is a (well-defined) operator on the set of such functions. Also, I have edited my answer in response to the OP's correction, so $P_1,Q_1$ are no longer even mentioned in the answer. – Cameron Buie Feb 12 '13 at 21:16
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If (as both the question title and initial sentence suggest) you are struggling with truth tables more generally, read a good introductory logic book. To help you with this sort of thing is exactly what such books are there for. If one doesn't work for you, try another. But you'll find patient explanations (more extended and careful that we can give here).

For example, you could look at Paul Teller's A Modern Logic Primer. Download the first four chapters from here: http://tellerprimer.ucdavis.edu/pdf/

Or you could look at my Introduction to Formal Logic or Sam Guttenplan's The Languages of Logic or Paul Tomasi's Logic (to mention just a few other books which aim to be particularly accessible at this introductory, and are thought to succeed).

Peter Smith
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