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)(Fx ⊃ E!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!x ⊃ Fx) is true if and only if every instance (E!a ⊃ Fa) 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.