# Are these two logical statements equivalent?

The question is:

Translate the following into logical notation:

Nobody is despised who can manage a crocodile.

Given that:

• $D(x)$ = "$x$ is despised"
• $M(x)$ = "$x$ can manage a crocodile"

It states the answer as being $\forall x (M(x) \rightarrow \neg D(x))$. But would it also be acceptable to say:

$\neg \exists x (D(x) \land M(x))$

• Why do we need $B(x)$ and $I(x)$? Jan 3 '18 at 12:45
• @NathanaelSkrepek Sorry about that, there were 2 different questions that I didn't include (because I got the same answer as is shown on the marking scheme). Jan 3 '18 at 12:55
• @Shannon Nice to see you use Mathjax! :) Jan 3 '18 at 17:13
• @Bram28 It is great, thank you again for showing me that tutorial! Jan 3 '18 at 17:29
• @Shannon You're welcome! :) Jan 3 '18 at 18:15

Yes. They are equivalent.

$\forall x (M(x) \rightarrow \neg D(x)) \\ \quad \Leftrightarrow \neg\exists x \neg(M(x) \rightarrow \neg D(x)) \\ \quad \Leftrightarrow \neg\exists x (M(x) \land \neg\neg D(x))\\ \quad \Leftrightarrow \neg\exists x (M(x) \land D(x))\\ \quad \Leftrightarrow \neg\exists x (D(x) \land M(x))$

The first equivalence comes from the familiar relation between the two quantifiers; the other equivalences should all look compelling given what you know about about e.g. the equivalence of $\neg(A \to B)$ and $(A \land \neg B)$.

(Careful: what I've just said is supposed to be motivational, give you a sense of why the equivalences hold. A proof in e.g. a natural deduction system will be more complicated since you can't apply connective rules directly to the innards of a wff inside the scope of a quantifier!)

Note that $$P \to Q \equiv \neg P \lor Q \tag 1$$ Also, note that $$\neg \exists x \, P (x) \equiv \forall x \neg P (x)$$

Thus, your books expression can be expressed as: $$\forall x (\neg D(x) \lor \neg M (x)) = \forall x (M (x) \to \neg D (x))$$ the same as your expression.

Nobody is despised who can manage a crocodile

Given that:

D(x) = "x is despised"

M(x) = "x can manage a crocodile"

It states the answer as being $$\forall x (M(x) \rightarrow \neg D(x))$$

But would it also be acceptable to say $$\neg \exists x (D(x) \land M(x))$$

I would say that it would be a more direct translation, since it maintains the structure of the sentence being translated.   Vis: "There does not exist somebody who is despised and can manage a crocodile."

They are, however, both equivalent in first order predicate calculus, and some lecturers will prefer you to avoid leaving expressions with negated quantifiers as a matter of style.   So $$\forall x~(M(x)\to\neg D(x))$$ ie "Anybody who can manage a crocodile is not despised," is also viable.

Though so is the contraposition: $$\forall x~(D(x)\to\neg M(x))$$ ie "Anybody who is despised can not manage a crocodile."

Well, anyway, what ever final form is preferred by your lecturer, it never hurts to show your working in an exam or assignment.   (Indeed it is preferred that you do so.)

One nice way to see the two statements are equivalent is by using Venn Diagrams.

We can represent the sentence $\forall x (M(x) \rightarrow \neg D(x))$ in a Venn diagram as follows: Explanation: When in a Venn diagram an area is shaded, that means that there do not exist any objects there. So, by shading the area that is both inside the '$M$' circle and inside the '$D$' circle, we are forcing that anything that is inside the '$M$' has to be outside the '$D$', and hence anything that is an '$M$' cannot be a '$D$' ... which is exactly what we want.

Now, compare this to the Venn Diagram for $\exists x (D(x) \land M(x))$: This time, we put an 'X' in the intersection, indicating that there is at least one object there, capturing $\exists x (D(x) \land M(x))$

Now, if you compare the two Venn Diagrams, you'll notice that they are saying the exact opposite: one is saying that there is not a thing in the intersection, while the other is saying that is is a thing in the intersection. Thus, one is the negation of the other. And thus:

$\forall x (M(x) \rightarrow \neg D(x)) \Leftrightarrow \neg \exists x (D(x) \land M(x))$