# Translating to symbolic logic and back

Yes, it is that time of the semester when all of my classes are having exams.  It seemed that the last posts going over the practice exam problems for Calculus were helpful, so I will continue to make such posts around exam time.  For this series of posts, I will be working through the practice exam for the exam my introduction to proofs class will take later this week.

Our goal in this post is to start with a quotation from Lewis Carroll.  We will then convert this into symbolic logic.  After that, we will find the negation of the statement and convert this into an English sentence.  If you would like to see more about Lewis Carroll’s use of logic, please view the post Alice in Logicland.  There are also many more examples in his book Symbolic Logic.

Everyone who is sane can do logic.  None of your sons can do logic. -Lewis Carroll in Symbolic Logic

You can also view the video of this solution on YouTube.

### Translating to symbolic logic

In order to turn this into a statement using symbolic logic, the first thing I want to do is to define any variables within the statement.  Note that, the statement will apply to people, so I will define $$p$$ as a person.

We now want to state any simple statements that are given.  If we this, we have $$q_{1}(p)=$$”p is sane,” $$q_{2}(p)=$$”p can do logic,” $$q_{3}(p)=$$”p is your son.”  When choosing these statements, we want to make sure that we cover all things stating in the quote.  Furthermore, we want to ensure that there are no connectives within these statements.  If there are any, we want to break the statements up further until we can no longer do so.

Now that we have our simple statements, we need to combine them with connectives.  In the first sentence, the everyone, is telling us that we will have the quantifier for all, $$\forall$$.  Now, we have everyone who is sane can do logic.  When trying to determine how this are connected, note that we are saying that if a person is sane they can do logic.  Therefore, we would connect these with an if then.  That is, the first sentence is $$\forall p, q_{1}(p) \rightarrow q_{2}(p)$$.

As we look at the second sentence, we note the none.  This is going to give us both a negation and a for all.  Furthermore, when we finish the sentence we see the same thing we did in the first one.  That is, we will again have the conditional if then.  Combining these, we see $$\forall p, q_{3}(p) \rightarrow \sim q_{2}(p)$$.

Since both of the sentences are given, we need both to be true in order for the whole statement to be true.  That is, we have to connect these with an and.  Therefore, we have that the quote is equivalent to $(\forall p, q_{1}(p) \rightarrow q_{2}(p)) \wedge (\forall p, q_{3}(p) \rightarrow \sim q_{2}(p)).$

Since we will have to negate this, I will actually write this using the equivalency we have for statements.  That is, I will replace $$r \rightarrow s$$ with $$\sim r \vee s$$ and combine when I can.  I then get,
\begin{align*}
&(\forall p, q_{1}(p) \rightarrow q_{2}(p)) \wedge (\forall p, q_{3}(p) \rightarrow q_{2}(p)) \\
&=(\forall p, \sim q_{1}(p) \vee q_{2}(p)) \wedge (\forall p, \sim q_{3}(p) \vee \sim q_{2}(p)) \\
&=\forall p, (\sim q_{1}(p) \vee q_{2}(p)) \wedge (\sim q_{3}(p) \vee \sim q_{2}(p)) \\
&=\forall p, (\sim q_{1}(p) \vee q_{2}(p)) \wedge \sim q_{3}(p)) \vee ((\sim q_{1}(p) \vee q_{2}(p)) \wedge \sim q_{2}(p)) \\
&=\forall p, (\sim q_{1}(p) \wedge \sim q_{3}(p)) \vee (q_{2}(p)) \wedge \sim q_{3}(p)) \\
&\vee ((\sim q_{1}(p) \wedge \sim q_{2}(p)) \vee (q_{2}(p) \wedge \sim q_{2}(p)) \\
&=\forall p, (\sim q_{1}(p) \wedge \sim q_{3}(p)) \vee (q_{2}(p)) \wedge \sim q_{3}(p))\vee (\sim q_{1}(p) \wedge \sim q_{2}(p))
\end{align*}
Here we ignore the last portion because $$p \vee \sim p$$ is a contradiction. Since the true only needs one true, we can leave this portion off.

### Negating

We can now negate this statement.  In order to do this, we recall that we can negate for all, by turning them into there exists and negating the remaining statement.  We can also negate ands by turning these to ors and negating the remaining statements.  We would follow a similar process going in the opposite direction as well.  We therefore have that
\begin{align*}
&\sim(\forall p, (\sim q_{1}(p) \wedge \sim q_{3}(p)) \vee (q_{2}(p)) \wedge \sim q_{3}(p))\vee ((\sim q_{1}(p) \wedge \sim q_{2}(p)))\\
&=\exists p, \sim ((\sim q_{1}(p) \wedge \sim q_{3}(p)) \vee (q_{2}(p)) \wedge \sim q_{3}(p))\vee ((\sim q_{1}(p) \wedge \sim q_{2}(p))) \\
&=\exists p, \sim (\sim q_{1}(p) \wedge \sim q_{3}(p)) \wedge \sim (q_{2}(p)) \wedge \sim q_{3}(p)) \wedge \sim (\sim q_{1}(p) \wedge \sim q_{2}(p)) \\
&=\exists p, (\sim \sim q_{1}(p) \vee \sim \sim q_{3}(p) )\wedge (\sim q_{2}(p) \vee \sim \sim q_{3}(p)) \wedge (\sim \sim q_{1}(p) \vee \sim \sim q_{2}(p)) \\
&=\exists p, (q_{1}(p) \vee q_{3}(p)) \wedge (\sim q_{2}(p) \vee q_{3}(p)) \wedge (q_{1}(p) \vee q_{2}(p)).
\end{align*}

### Translating to English

We now have the symbolic negation and we have simplified far enough since we have distributed the negation out.  In order to convert this back to English, we begin by reading the statement as given.  This states, “There exists a p such that the person (is sane or is  your son) and  (cannot do logic or is your son) and (is sane or can do logic).”  Trying to combine this into a more colloquial sentence, we would arrive at “You have a son that is sane or can do logic, or there is someone that is sane and cannot do logic.”

## Conclusion

While it was not asked for in the problem, it would be worth converting the simplified version of the original statement back into English.  If we do this, we arrive at “None of your sons are sane or can do logic and if anyone else can’t do logic, they are insane.”  It is rather interesting to see such an insult thrown at a person, since such a person would likely not be able to determine what you are saying without a keen grasp on logic.

If you found the post helpful or entertaining, please like it below and share it with anyone else that may be looking at set theory or logic.  You can also find the rest of the solutions to this practice exam here.

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