Can a Dog Conclude It's a Cat Using Logic?

[bobby2k]

Homework Statement

This question may sound weird, but please bear with me.

Let's say that you are a dog, and you think to yourself.
"All cats have four legs, I have four legs, therefore I am a cat."
Obciously this is wrong because even though all cats have four legs, there are more creatures that have four legs.

But how would we write this in terms of logic? That is in terms of statements and the connectives → ,\wedge,\vee?

The Attempt at a Solution

My attempt is that I define a predicte:
F(x) = "x have four legs".
The predicate C(x) is "x is a cat".

Then I say that statement A is:
\forallx[C(x)→ F(x)]
statement B is:
F(I), that is "I have four legs".
Statement D is:
C(I) "I am a cat"

Now how can I see technically that
A\wedgeB → D is false?
This last step I can't get to.

 

[verty]

You can't see that it is false because you can't prove the theorem (or do not have the axiom) ~C(I). You have not included the knowledge that I am a dog and no dog is a cat. If you add these, you can prove ~C(I).

 

[bobby2k]

You can't see that it is false because you can't prove the theorem (or do not have the axiom) ~C(I). You have not included the knowledge that I am a dog and no dog is a cat. If you add these, you can prove ~C(I).

Ok, is this considered formally correct?
F(x) is "x have four legs".
C(x) is "x is a cat"
D(X) is "x is a dog"

statement A:\forallx[C(x)→ F(x)] is TRUE
statement B: F(I), that is "I have four legs". is TRUE
statement E: \forallx[D(X)→\negC(x)] is TRUE, this is what you said.

Now the proposition is:
A\wedgeB\wedgeD(I) → C(I)

Now I am a little stuck. I see that statement E says that D(I)→\negC(I), but what happens when we do not have D(I), but A\wedgeB\wedgeD(I)?
I guess it may be stupid, but can we just say that if we have statements Q, P, H and
Q→\negP
then: Q\wedgeH →\negP?

 

[HallsofIvy]

Your "argument" would be
If P then Q
Q
Therefore P.

where P= "is a cat", Q= "has four legs".

This is called "affirming the consequent".