First, let's formulate these statements. If $S$ is the set of all squirrels, $M$ is the set of all monkeys, $R$ is the set of all rodents and $a$ is an animal that could be in any one of those sets, we get $\begin{aligned} a\in S &\implies a\in R\\ a\in R &\implies a\in M. \end{aligned}$ From which we can immediately conclude that $a\in S \implies a\in M$ . So every squirrel has to be a monkey. Note though, that these are implications, not equivalences. If they were equivalences (meaning $a\in S\implies a\in M\;\wedge\; a\in M\implies a\in S$ ), then we could conclude that all monkeys are squirrels. Look at this Venn-Diagram to see what I mean:

You can immediately see that there are monkeys that are neither squirrel nor rodent.

$\text{All squirrels are rodents} \Rightarrow \text{Squirrels ∈ Rodents}$

$\text{All rodents are monkeys} \Rightarrow \text{Rodents ∈ Monkeys}$

Hence, the answer is $\boxed{\text{No}} \text{, since the monkeys in the red-and-blue area aren't squirrels, and the ones in blue aren't even rodents.}$

This is a classic syllogism where , through transitivity, All p's are q's and All q's are r's can be combined into All p's are r's. In this case, this leads to : All p's are r's. But this does not imply the converse, that is all monkeys are squirrels. In fact, even the converse of the second statement is not true in general, i.e. All monkeys are rodents is not true in general. It is true that some monkeys are rodents and some rodents are squirrels but they may be disjoint sets.

Let's call the set of squirrels $S$ , the set of rodents $R$ , and the set of monkeys $M$ . From the problem, we have $S∈R$ , and $R∈M$ . From this, we can construct a Venn diagram; I can leave a space between the circles because the problem didn't state that all rodents are squirrels, and that all monkeys are rodent. From the Venn diagram, we can easily see that there is some space between the rodent's circle and the monkey's circle, which must mean that there is some creature that is a monkey, but not a rodent or a squirrel, so therefore, the answer is $\boxed{No}$

Let $S,R,M$ notate the set for squirrels, rodents and monkeys. Then we have $S\in R, R\in M. ~~~\therefore S\in M.$ I can therefore conclude all squirrels are monkeys (what?!). But the question asks whether all monkeys are squirrels, which is whether $M\in S$ .
Since we have $S\in M$ already, either $M\in S$ is true and $M=S$ or $M\in S$ is false and $M\neq S$ .
We cannot be sure, so the answer is NO. This Venn diagram may help :)

Terminologies in Set Theory

Elements : Objects belonging to a set

Subset : If all elements from a set (say A) are also element of another set (say B), then A is subset of B. It is denoted as

A ⊆ B

1. All $\color{#3D99F6}{Squirrels}$ are $\color{#CEBB00}{Rodents}$ . From this we can conclude that the set of all $\color{#3D99F6}{Squirrels}$ is a subset of the set of all $\color{#CEBB00}{Rodents}$ i.e. $\color{#3D99F6}{Squirrels}$ ⊆ $\color{#CEBB00}{Rodents}$ .

2. All $\color{#CEBB00}{Rodents}$ are $\color{#20A900}{Monkeys}$ . From this we can conclude that the set of all $\color{#CEBB00}{Rodents}$ is a subset of the set of all $\color{#20A900}{Monkeys}$ i.e. $\color{#CEBB00}{Rodents}$ ⊆ $\color{#20A900}{Monkeys}$ .

Conclusion

$\color{#3D99F6}{Squirrels}$ ⊆ $\color{#CEBB00}{Rodents}$ ⊆ $\color{#20A900}{Monkeys}$ (this is represented by the Venn Diagram).

From this it is clear that, with given information, though all $\color{#3D99F6}{Squirrels}$ are $\color{#20A900}{Monkeys}$ , all $\color{#20A900}{Monkeys}$ are not $\color{#3D99F6}{Squirrels}$ . (Considering all sets are not equal)

E.g. $\color{#20A900}{Monkey}$ represented by point A on Venn Diagram is not a $\color{#3D99F6}{Squirrel}$ .

First let's look at an analogy:

1. All Natural Numbers(N) are Whole Numbers(W). This does not mean that all whole numbers are natural numbers. For example, 0 is a whole number but not a natural number.

2. All Whole Numbers(W) are Integers(Z). Similarly all negative integers are not whole numbers.

So, all natural numbers are integers but all integers are not natural numbers(-2,-4,0 etc.) $\textcolor{#FFFFFF}{\text{Venn diagram representation}}$

Similarly, we infer from the information given that all squirrels are monkeys, but monkeys are not squirrels. all monkeys are not squirrels

Image Since all squirrels are only part of the group of monkeys, and there are other monkeys which are not squirrels, the answer is $\boxed{No}$ .

$\red{S=\{x:x\space is\space a\space squirrel\}}$ $\blue{R=\{x:x\space is\space a\space rodent\}}$ $M=\{x:x\space is\space a\space monkey\}$ $\red{p:All\space squirrels\space are\space rodents.}$ $\blue{q:All\space rodents\space are\space monkeys}$ $p\Rightarrow S\subseteq R$ $\red{q\Rightarrow R\subseteq M}$ $\blue{\Rightarrow p \space and\space q\Rightarrow S\subseteq M}$ $\cancel{\Rightarrow} (x\in M\Rightarrow x\in S)$ $\Rightarrow \boxed{not \space all\space monkeys\space are\space squirrel}$ $Hence\space the\space answer\space is\space \boxed{NO}$

First from set theory we can call all squirrels as rodents, and all rodents as monkeys. Now the population of rodents(includes squirrels) can't exceed the population of monkeys because rodents is subset of monkeys (A subset value never exceeds that of its set), a similar statement can be said for squirrels (subset) and rodents (set)

There will be 4 cases because there nothing mentioned about the population.

• When population of each animal is different.
• When only population of rodents and squirrels are equal.
• When all 3 have same population.
• When only population of rodents and monkeys are equal.

These are respective Venn Diagrams.

From the Venn diagrams, we can see that we can't necessarily tell that all monkeys are squirrels. Therefore ans is $\red{Nooooo.......}$

Given $Squirrels \subset rodents$ and $rodents \subset monkeys$ We can conclude that $squirrels \subset monkeys$ because subset is a transitive relationship, but we can not conclude that $monkeys \subset squirrels$ . Although the given statements do not rule out all three sets being exactly the same, we can not conclude that that is the case.

All squirrels are rodents

This statement tells us that every squirrel is a rodent.

This does not necessarily mean that every rodent is a squirrel. Squirrels are only one of the things that are rodents. For example, all rats are rodents, and so are the squirrels. It doesn’t mean that squirrels are rats.

Here is the diagram:

Rats are now not part of this anymore, we move on to the next statement.

All rodents are monkeys

If all rodents are monkeys, this makes the amount of monkeys equal or more than the amount of rodents. The statements do not say that all monkeys are rodents.

Squirrels = or < Rodents = or < Monkeys

This also suggests that all squirrels are also monkeys. Why that is, I don’t know, but that’s the idea.

$Not$ $all$ $monkeys$ $are$ $squirrels$

(Irl, no monkeys are squirrels, and no squirrels are monkeys. I think)

Let R be the set of all rodents,

S be the set of all squirrels and

T be the set of all monkeys

According to the first statement,

All Squirrels are rodents

=> Squirrels lie in the power set of rodents I.e. squirrels are a part of the set of all rodents i.e,

S ∈ R

Also,since S is a part of R ,the opposite( R ∈ S ) is not true.

Our next statement says,

All rodents are monkeys.

=> Rodents lie in the power set of monkeys I.e. rodents are a part of the set of all monkeys i.e,

R ∈ T

Again,the opposite isn't true

Venn Diagram

Hence,

The biggest set is that of monkeys' in which lies the set of rodents (smaller than that of monkeys' but bigger than squirrels') in which lies the set of squirrels

As you can clearly SEE , ALL MONKEYS ARE NOT SQUIRRELS ……ᘛ⁐̤ᕐᐷ

Squirrels are in the set of rodents and rodents are in the set of monkeys but monkeys are an independent set being inside no-one As you can see inside the venn diagram,

so in this horrible drawing I express that U set are monkeys and all the compounds in that set are squirrels and rodents, so the logic behind that to say that not all monkeys are squirrels is the following example: Imagine that U set is the universe and in the other sets they are galaxies, then planets, so if I say that all the universes are in planets, it is wrong because all the planets are in the universe (and we know that the universe is not composed only of planets so it is not true) (we can define them as "some"). The explanation is clear that the universe set is not necessarily the set of planets. The same goes for monkeys, monkeys are not necessarily squirrels. (The interesting thing about set theory is that we can never express everything is nothing, just in our mind).

Let $S$ denote the set of all squirrels, $R$ denote the set of all rodents and $M$ denote the set of all monkeys, we're given that,

All Squirrels are rodents

All rodents are monkeys i.e,

$S⊆R⊆ M$

Which means that $M$ is the Universal Set of which $R$ is a subset and $S$ again is a subset of $R$ implying also of $M$ . For the required case to be true, i.e one that says "All monkeys are squirrels" we need $M\subset S$ , but since already $S\subset M$ , for the required condition to be true $S=M$ .

From the given two statements two conditions arise,

• $R$ is a proper subset of $M$ , $R<M$ , not all elements of $M$ are in $R$

Since $S$ is a subset of $R$ , $S$ will always be less than $M$ too.

• $R$ is an improper subset of $M$ , $R=M$ , every element of $M$ is in $R$

Since $S$ is a subset of $R$ , either it is a proper subset $S<(R=M)$ or an improper one $S=(R=M)$ .

In the above conditions, only one, $S=(R=M)$ or $S=M$ supports the required case, since it is not necessarily true that $S=M$ , it is not necessarily true that "All monkeys are squirrels" .

For the sake of generality we've not considered trivial improper subsets, or null subsets, though if $R$ is null, $S$ is also null which doesn't support the required case either.

Most Common Solution -

All squirrels are rodents. This means that every last squirrel on this planet is a rodent. This doesn't necessarily mean that all rodents are squirrels, here's how -

We can say that all apples are fruits. This doesn't mean all fruits are apples. Mangoes and bananas aren't apples. So, if all A are B, that doesn't mean all B are A.

So, not all rodents are squirrels.

Now, the second statement states, 'All rodents are monkeys' and again same thing, all rodents are monkeys, but not all monkeys are rodents.

Not all monkeys are rodents, which means that there are some monkeys that are not rodents. These monkeys aren't squirrels either, because all squirrels are rodents. Thus, the answer is $\boxed{\text{No}}$

Maybe being judge means you have to solve the problem first all the time?