Logic expression into english

[Miike012]

Is the statement "if you are a comedian then you are funny." equivalent to what is defined in part (a)?

 

[verty]

"You" is singular because "a comedian" is singular, so it is applying to a single person. One single person is funny if they are a comedian.

Hint: the correct translation is very short.

 

[CompuChip]

"You" in this case means "a person", i.e. "any person" (hint: any ${}\approx\forall$ ;-))

 

[Miike012]

The statements says all comedians are funny. The reason why I am asking the question is because to me the symbolic expression in the paint document which expresses that "All comedians are funny" implies that if you are a comedian then you must be funny.

So what I am saying, that "all comedians are funny" and "if you are a comedian then you are funny" are equivalent.

After all the logical expression is referring to all people in that world that are comedians which is exemplified in the sentence "if you are a comedian then you are funny."

That is what I am thinking and to me my logic makes sense but I am obviously work.

 

[Miike012]

Here is another explanation of what I am thinking.

First assume that all comedians are funny.

Now Let the set P consist of all people and let C consist of n elements, none of which are equal to each other. If
The set C consists of all comedians in P, then there must only be n people that are comedians.

Now I will construct a set c such that c = {c₁,c₂,...,c_m}. I will define set c to contain comedians therefore if c_m is a person and a comedian then it will belong to the set c.
Because c is a subset of P, m must be equal to n, or c must be comprised of n comedians. I can conclude that c and C are equal.

where C denotes "all comedians are funny."
and c denotes "if you are a comedian then you are funny."

I think I know where I am getting confused. Because I am constructing the set of comedians in my mind I am pretending that I am asking everyone the question "Are you a comedian" then they answer yes or no. If yes then I add them to the group of "Funny" if no then I ignore them.
However what I am doing is constructing the set that already "exists". What I should do is not construct the set but already assume that the set exists.
I bet none of what I just said makes any sense.

 

[CompuChip]

I think you were getting a bit carried away in your second post, so let me reply to the first one instead :)

The statements says all comedians are funny. The reason why I am asking the question is because to me the symbolic expression in the paint document which expresses that "All comedians are funny" implies that if you are a comedian then you must be funny.

So what I am saying, that "all comedians are funny" and "if you are a comedian then you are funny" are equivalent.

Correct. This is how in mathematical logic we express a statement like "all X are Y". You cannot express this directly, because you would have to say something like "Let $C$ be the set of all comedians. Then $\forall x \in C, F(x)$. Unfortunately, in logic, a for-all symbol always runs over the entire universe - in this case, all people, so once you write $\forall x$ you imply $\forall x \in X$ and the qualification over $C$ is invalid.

Instead, you know that if you find a comedian, then (s)he must be funny. This is one of the things that the implication $C(x) \implies F(x)$ expresses. Also, note that the statement does not say anything about what happens when x is not a comedian. We are not saying that all non-comedians are funny too, or aren't funny. Otherwise, you would need to express this additional information.
Just for your understanding, it may be helpful to compare the following statements and trying to translate them to a logical formula:
(1) "All comedians are funny." (the original statement)
(2) "All comedians, and comedians only, are funny."
(3) "All funny people are comedians."

If you have learned about truth tables, you may want to try drawing those too and look at the differences:

C(x)  F(x)   (1): C(x)->F(x)   (2): ...    (3): ...
  T      T              ...
  T      F
  F      T
  F      F