MHB Valid Arguments: Sets Explained

[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.

 

[Ackbach]

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?

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

This is all correct.

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?

 

[ertagon2]

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?
Can you please explain 6.(3)
Why is it $\therefore$p$\implies$r rather than p$\therefore$r <--this is what I meant

 

[Ackbach]

So I did answer everything correctly?

Yes, I think so.

Can you please explain 6.(3)
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?