*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.