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 X X

2. The argument is invalid

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

The other branches are open. We get an interpretation

D = {a}

 I(F) I(E!) a 1 0

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 0 1 1

And the conclusion is false, as we have

 E!a & Fa 0 0 1

Therefore, the argument is invalid.

3. The argument is invalid.

 (∀ x)(Gx ⊃ E!x) \a E!a ~~ Fa Fa / \ ~ E!a Ga ⊃ E!a X / \ ~ Ga E!a

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

D = {a}

 I(F) I(G) I(E!) a 1 0 0

makes the premises true, as we have

 E!a ⊃ (Fa ⊃ E!a) 0 1 1 0 0

and

 E!a 1

and the conclusion

 ~ Fa 0 1

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)(Fx ⊃ E!x) \a ~ E!a ~~ Fa Fa / \ ~ E!a Fa ⊃ E!a / \ ~ Fa E!a X X

We get the following counterexample from the leftmost branch:

D = {a}

 I(F) I(E!) a 1 0

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.