# How to Negate These Statements Containing Logical Connectives?

I have a few questions that I am working on, that I supposedly answered incorrectly.

I have the following statements that I am charged to express in symbolic form:

$f =$ you are a full-time student;
$o=$ you are over 21;
$r=$ you are a resident of the state; and
$e=$ you are eligible for financial aid.

b) You are neither a resident nor are you over 21

My answer: $\neg r \lor \neg o$

d) To be eligible for financial aid it is necessary to be a full-time student

My answer: $f \rightarrow e$

Here is another, but similar, question, where I am suppose to express the statement in if-then form:

a) A necessary condition for this computer program to be correct is that it does not produce error messages during translation.

My answer: If a computer program does not produce any error messages during translation, then the computer program is correct.

The next question is one where I am to negate a few statements:

a) John is not wealthy but he is healthy and wise.

My answer: John is wealthy but he is neither healthy nor wise.

b) Stocks are increasing but interest rates are not steady nor are they low.

My answer: Stocks are not increasing but interest rates are steady and they are low.

I was wondering if someone could possibly tell me if these are truly incorrect, because feel as though I answered them properly.

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No they are all wrong... Take for example the last one. The opposite of "nor" is not "and" but "or". For example: the opposite of stating that "the cake is not sweet nor big", is the cake is sweet or big. –  PML Mar 13 '13 at 19:18
You can check the wiki article on De Morgan's rules. Specially the "formal proof" section if you have doubts. –  PML Mar 13 '13 at 19:24

For your first (b): "You are neither a resident nor are you over 21"

This is equivalent to saying "it is not the case that you are a resident, and it is not the case that you are over 21".

$$\text{"Neither r nor o"}\quad \iff \;\;\lnot(r \lor o) \quad \equiv \quad \lnot r \land \lnot o$$

For part (d): See the following link about different ways to say "if-then"

You'll see that $;\;p \rightarrow q\;\;$ is equivalent to either of the following:

• ="$p$ is a sufficient condition for $q$",

• "$q$ is a necessary condition for $p$".

That is, in the implication $\;p \rightarrow q:\;$: $\,p\,$ is the sufficient condition for $q$, and $\,q\,$ is the necessary condition for $p$.

Another way to look at the second bullet point: q is a necessary condition for p, means essentially, that if not q, then not p $\equiv \lnot q \rightarrow \lnot p\equiv p \rightarrow q$

Sample negation: we'll negate the proposition

"John is not wealthy but he is healthy and wise."

Let's first translate to $r$ = John is wealthy (r for rich); $h$ = John is healthy; $w$ = John is wise.

That the given statement is $$\lnot r \land h \land w\tag{1}$$

Negating this gives us $$\lnot(\lnot r \land h \land w) \equiv r \lor \lnot h \lor \lnot w\equiv r \lor \lnot(r \land w)\tag{\lnot (1)}$$ by DeMorgan's.

So translating $\lnot(1)$ gives us:

"Either John is rich, or John is not healthy or John is not wise." Equivalently, "John is rich or John is not both healthy and wise."

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Yes, I believe so. Is there some relationship between word either in English and using a parenthesis in propositional calculus? –  Mack Mar 13 '13 at 20:06
Either precedes or: either a or b $\iff a \lor b$. Think of "neither-nor" as the negation of "either-or": i.e. as $\lnot( a\lor b) \equiv \lnot a \land \lnot b.$ Neither $a$ nor $b$ –  amWhy Mar 13 '13 at 20:12
So, neither $a$ nor $b$ corresponds to $\neg a \wedge b$? –  Mack Mar 13 '13 at 20:17
Neither-nor corresponds to the the negation of $a\lor b:\, = \lnot(a \lor b) \equiv \lnot a \land \lnot b$, by DeMorgan's. –  amWhy Mar 13 '13 at 20:18
Oh, I see. Thank you! –  Mack Mar 13 '13 at 20:23

(b) One correct answer is $\lnot r \land \lnot o$. Both $r$ and $o$ must be false. An alternative correct expression would be $\lnot(r\lor o)$.

(d) One correct answer is $e \to f$. Your version says that if you are a full-time student, then you are eligible. There might be other conditions, such as being a resident. The English version said it was necessary to be a full time student. It did not say that being full-time is sufficient.

Your answer to the computer program question is not correct, and essentially repeats the mistake in (d). In English, if should be "If this computer program is correct, then $\dots$"

The answer to the John question is not correct. Essentially the original statement is not wealthy and healthy and wise. So the statement says all of the assertions are true. The negation should say that at least one of the parts fails. So it should say wealthy or not healthy or not wise.

The stock question should be done like the solution we gave for the John question.

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@amWhy: I hope that you will not mind this post as an Answer (and not a comment), granted that I am not permitted to learn about the rejectors of my edits.

Would it be more helpful and less confounding to accentuate and promulgate the fact that your last sentence relates more to the middle expression (and NOT the rightmost one) in $(\lnot(1))$?

I tried to accomplish this with green colour but do not understand the rejections for my edits. If it had been approved, the latter part of your Answer would have become:

Sample negation: we'll negate the proposition

"John is not wealthy but he is healthy and wise."

Let's first translate to $r$ = John is wealthy (r for rich); $h$ = John is healthy; $w$ = John is wise.

That the given statement is $$\lnot r \land h \land w\tag{1}$$

Negating this gives us $$\lnot(\lnot r \land h \land w) \equiv \color{green}{r \lor \lnot h \lor \lnot w}\equiv r \lor \lnot(r \land w)\tag{\lnot (1)}$$ by DeMorgan's.

So translating $\lnot(1)$ gives us:

"Either John is rich, or John is not healthy or John is not wise." Equivalently, the green says: "John is rich or John is not both healthy and wise."

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