# Valid Arguments: Sets Explained

• MHB
• ertagon2
In summary, a valid argument in set theory is one where the conclusion logically follows from the premises. To determine the validity of an argument, one must identify the premises and the conclusion and evaluate whether the conclusion logically follows from the premises. The difference between a valid and an invalid argument is that a valid argument has a conclusion that logically follows from the premises, while an invalid argument does not. A valid argument can have false premises, but it may not be sound. Sets can be used in valid arguments as a way to organize and categorize information, make logical statements, and support arguments. They also allow for working with complex relationships and making more precise arguments.
ertagon2
View attachment 7799
is this right ?6(1).
p$\implies$q,
$\lnot$q$\therefore\lnot$p,
Modus tollens

6(2).
p$\implies$q,
q$\implies$r,
q$\therefore$r,
Transitivity

6(3).
p$\implies$q,
q$\therefore$p.
Converse fallacy

7. I have no idea what I'm doing please explain.

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ertagon2 said:
is this right ?6(1).
p$\implies$q,
$\lnot$q$\therefore\lnot$p,
Modus tollens

6(2).
p$\implies$q,
q$\implies$r,
q$\therefore$r,
Transitivity

I think you mean $\therefore \; p\implies r$, right?

ertagon2 said:
6(3).
p$\implies$q,
q$\therefore$p.
Converse fallacy

This is all correct.

ertagon2 said:
7. I have no idea what I'm doing please explain.

The subset $\subset$ relation works like this: $A\subset B$ if and only if every element of $A$ is an element of $B$. So, for 7.A., look at every element of the set $\{1,2\}$, and see if they are in $A$ (they are). 7.B. is trickier. The relation "is an element of", $\in$, works differently from the $\subset$ relation. Sets can be elements of other sets, or they can be subsets of other sets. Neither implies the other. In this case, because $A=\{1,2,\{1,2\}\}$, and you have that nested $\{1,2\}$ sitting inside, then it follows that $\{1,2\}$ is an element of $A$. For something to be an element of another set, it's got to show up at the highest level, separated from other elements by commas. So, for example, if you define the set $C=\{1,2,\{3,4\}\}$, then the elements of $C$ are $1$, $2$, and $\{3,4\}$. Neither $3$ nor $4$ are elements of $C$. Does this help? Can you see why 7.C. is true and 7.D. is false?

Ackbach said:
I think you mean $\therefore \; p\implies r$, right?
This is all correct.
The subset $\subset$ relation works like this: $A\subset B$ if and only if every element of $A$ is an element of $B$. So, for 7.A., look at every element of the set $\{1,2\}$, and see if they are in $A$ (they are). 7.B. is trickier. The relation "is an element of", $\in$, works differently from the $\subset$ relation. Sets can be elements of other sets, or they can be subsets of other sets. Neither implies the other. In this case, because $A=\{1,2,\{1,2\}\}$, and you have that nested $\{1,2\}$ sitting inside, then it follows that $\{1,2\}$ is an element of $A$. For something to be an element of another set, it's got to show up at the highest level, separated from other elements by commas. So, for example, if you define the set $C=\{1,2,\{3,4\}\}$, then the elements of $C$ are $1$, $2$, and $\{3,4\}$. Neither $3$ nor $4$ are elements of $C$. Does this help? Can you see why 7.C. is true and 7.D. is false?

So I did answer everything correctly?
Why is it $\therefore$p$\implies$r rather than p$\therefore$r <--this is what I meant

ertagon2 said:
So I did answer everything correctly?

Yes, I think so.

ertagon2 said:
Why is it $\therefore$p$\implies$r rather than p$\therefore$r <--this is what I meant

The problem with writing $p\therefore r$ is that it is confusing. Someone looking at that might think you've concluded that $p$ is actually true, whereas it's only ever an assumption in the entire argument. Moreover, the word "therefore" or symbol $\therefore$ is meant to indicate that what follows is the conclusion. It's not usually meant to be used as part of the conclusion, or in the middle of it. Does that help?

## 1. What is a valid argument in set theory?

A valid argument in set theory is a series of statements where the conclusion follows logically from the premises. In other words, if the premises are true, then the conclusion must also be true. This is known as a "valid" argument.

## 2. How do you determine the validity of an argument?

To determine the validity of an argument, you must first identify the premises and the conclusion. Then, you must evaluate whether the conclusion logically follows from the premises. If it does, then the argument is considered valid.

## 3. What is the difference between a valid and an invalid argument?

A valid argument is one where the conclusion logically follows from the premises, while an invalid argument is one where the conclusion does not logically follow from the premises. In other words, an invalid argument can have true premises but still have a false conclusion.

## 4. Can a valid argument have false premises?

Yes, a valid argument can have false premises. As long as the conclusion follows logically from the premises, the argument is considered valid. However, if any of the premises are false, the argument may not be sound.

## 5. How can sets be used in valid arguments?

Sets can be used in valid arguments as a way to organize and categorize information. By defining sets and their elements, we can make logical statements about those sets and use them to support our arguments. Sets also allow us to work with complex relationships and make arguments that are more precise.

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