# Problem on Logic

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1. Jan 11, 2015

### 22990atinesh

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"

2. Jan 11, 2015

### Staff: Mentor

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.

3. Jan 11, 2015

### spinnaker

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.

4. Jan 13, 2015

### 22990atinesh

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.

5. Jan 13, 2015

### Staff: Mentor

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

6. Mar 25, 2015

### MLP

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.

Last edited: Mar 25, 2015
7. Mar 26, 2015

### LAZYANGEL

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.

8. Mar 31, 2015

### Apogee

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.

9. Mar 31, 2015

### micromass

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!

10. Apr 1, 2015

### WWGD

You may have a world without comedians. An actual argument would be to construct a world/ interpretation without comedians. The second statement says there is an x, so that _if_ x is a comedian, x is funny. But there may be no comedians. And, yes, if all comedians are funny and there are comedians, then there is a funny comedian. But there are formal rules for derivations in Predicate Logic you need to follow to rigorously show this. If you want a world without funny people, just take a bunch of copies of Adam Sandler.

Formally, you can construct a world/interpretation in which the first is true , but the second one is not, to show the two are not equivalent. For this, take a world with just one funny comedian and one non-funny one (Adam Sandler ) and the other world is one with two comedians that are both funny.

Last edited: Apr 1, 2015
11. Apr 8, 2015

### mac_alleb

First say is overall.
Second is exasperate.
So they mean differ.