Problem on Logic

Consider the predicates
C(x)=x is a comedian
F(x)=x is funny
where domain=All people in the world

Now consider the predicate logic

##1. \forall x [C(x) \rightarrow F(x)]##

##2. \exists x [C(x) \rightarrow F(x)]##

Doesn't both the above predicate logic refers to the same conclusion that "All Comedians are funny"

Mark44
Mentor
Consider the predicates
C(x)=x is a comedian
F(x)=x is funny
where domain=All people in the world

Now consider the predicate logic

##1. \forall x [C(x) \rightarrow F(x)]##

##2. \exists x [C(x) \rightarrow F(x)]##

Doesn't both the above predicate logic refers to the same conclusion that "All Comedians are funny"
No, they aren't both saying the same thing. The first one says that all comedians are funny, but the second one says only that at least one comedian is funny.

Agreed. Assume x means people, and C(x) means people that become comedians.

First statement reads, for all people that become Comedians, they become funny comedians.

Second statement reads, there exists someone who, when they become a comedian, is funny.

Agreed. Assume x means people, and C(x) means people that become comedians.

First statement reads, for all people that become Comedians, they become funny comedians.

Second statement reads, there exists someone who, when they become a comedian, is funny.

No, they aren't both saying the same thing. The first one says that all comedians are funny, but the second one says only that at least one comedian is funny.

Doesn't (2) says this
If there exits a person who is comedian then he is funny ##\implies## Among all people those who are comedian are funny ##\implies## all comedian are funny.

Mark44
Mentor
Doesn't (2) says this
If there exits a person who is comedian then he is funny ##\implies## Among all people those who are comedian are funny ##\implies## all comedian are funny.
No, the second statement doesn't mean this. All it says is that there is at least one commedian who is funny.

No, the second statement doesn't mean this. All it says is that there is at least one commedian who is funny.

I realize this is an older OP but since some of the postings contain misinformation, perhaps this reply is relevant.

The second statement DOES NOT mean there is at least one one comedian who is funny. The translation for 'There is at least one comedian that is funny' is

∃x[C(x) & F(x)]

∃x[C(x)→F(x)] is logically equivalent to ∃x[~C(x) v F(x)] which is made true in a domain where there is at least one non-comedian or funny person.

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I've modified it a bit:

##\forall x \: [C(x) \rightarrow F(x)],\: x \in \mathbb{Z}##
All comedians are funny.

##\exists x \: [C(x) \rightarrow F(x)],\: x \in \mathbb{R}##
Yet only some of them really are.

The first statement establishes that all comedians are funny. The second statement merely establishes that only one comedian need be funny. There could be millions of comedians that are not funny so long as at least one is funny. Therefore, the second statement would be satisfied, and many comedians would not be funny. This refutes the first statement. It is possible that the first and second statements are true, for the first statement implies the second statement. If all comedians are funny, then there exists at least one comedian that is funny. But the converse, if there exists one comedian that is funny, then all comedians are funny. That is the current flaw in your logic.

micromass
Staff Emeritus
Homework Helper
The second statement is not at all the same as "there is a funny comedian". In fact, if ##x## is not a comedian, then ##C(x)\rightarrow F(x)## is always true. So the second statement is always true if there is somebody who is not a comedian!

austinuni
WWGD