Logic Text Chapter 13 Solutions

Chapter 13: Some Things do not Exist

Basic

Question {13.1}

1. The tree closes, the argument is valid.

(∃ x)Fx / a
~(∃ x)(E!x & Fx) \a
|
E!a
Fa
/\
~ E!a~(E!a & Fa)
X /\
~ E!a~ Fa
XX

2. The argument is invalid

(∀ x)Fx \a
~(∃ x)Fx / a
/\
/\
~ E!aFa
/\/\
~ E!a~ Fa~ E!a~ Fa
*X

The other branches are open. We get an interpretation

D = {a}

I(F)I(E!)
a 10

from the branch with the asterisk. (Or Fa being false would be equally as appropriate, had we chosen a different branch)

The premise is true, as we have

E!a Fa
011

And the conclusion is false, as we have

E!a &Fa
001

Therefore, the argument is invalid.

3. The argument is invalid.

(∀ x)(GxE!x) \a
E!a
~~ Fa
Fa
/\
~ E!aGaE!a
X /\
~ GaE!a

The tree is open as the rightmost branch remains open. The interpretation is:

D = {a}

I(F)I(G)I(E!)
a 100

makes the premises true, as we have

E!a(FaE!a)
01100

and

E!a
1

and the conclusion

~Fa
01

is false, showing the argument to be invalid.

There is a mistake in the question the correct version should be:

(∀ x)(FxE!x), ~ E!a, therefore ~ Fa (see errata)

In this case the argument is invalid, the tree is

(∀ x)(FxE!x) \a
~ E!a
~~ Fa
Fa
/\
~ E!aFaE!a
/\
~ FaE!a
XX

We get the following counterexample from the leftmost branch:

D = {a}

I(F)I(E!)
a 10

Question {13.2}

In any interpretation (Ax)(E!xFx) is true if and only if every instance (E!aFa) is true (for each a in the domain). But this is true iff Fa is true for each a in the inner domain, and that happens iff (∀ x)Fx is true.

Similarly, (Sx)(E!x & Fx) is true if and only if some instance (E!a & Fa) is true (for some a in the domain), and this is true iff Fa is true for some a in the inner domain, which happens iff (∃ x)Fx is true.