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I understand most of the rules regarding bound variables, substitution instances etc. (or at least I think I do), but the issue is that I don't seem to be able to take it all the way. For instance, given the interpretation:

Domain: {1,2,3}
a: 1 b: 2 c: 3
F: {1,2}
G: {3}
H: {}
R: {(1,1), (2,1), (1,3), (3,3)}

How does one determine the truth value of the sentence: (Ey)(Ax)Rxy

I understand that in the substitution instance for 'a' we end up with (Ax)Rxa, but I don't know how to take it any farther. Essentially, my textbook explains the rules for getting to this point, but not what the next step is!

Do I end up having to simply evaluate (Ax)Rxa? So: Raa, Rba, Rca, making the substitution instance false since Rca is false and it's a universal quantifier? (And this holds true for the substitution instances for b and c as well, making the whole sentence false, since (Ex) needs at least one instance to be true?)

Another example: (-(Ez)Hz) -> {(Ax)[Fx v (Ey)(-Fy & Rxy)] -> (Az)Gz}

Then (substitution instance for a):
-Ha -> (Ax)[Fx v (Ey)(-Fy & Rxy)] -> (Az)Gz}
-Ha -> {[Fa v (Ey)(-Fy & Rxy)] -> (Az)Gz}
...and so on? Then, once the truth value for substitution instance 'a' has been established, repeat with substitution instance 'b' then 'c'? It seems extraordinarily unwieldy, especially seeing as I would then have to evaluate the "smallest" quantifier by a, b, c first, then do the same with the next largest, and etc.

My final question is regarding a sentence with the form: (Ex)Fx & (Ex)Gx

For substitution instance 'a', would I start with Fa & (Ex)Gx?

Then:
Fa & Ga = F
Fa & Gb = F
Fa & Gc = T

Making the whole sentence true? (due to only needing one instance to be true since it's an existential quantifier). If the sentence was instead (Ax)Fx & (Ex)Gx, I would need to also run the substitution instances for b and c as well (so Fb & Ga, Fb & Gb, etc.).

I've been trying to figure this out for the last couple of days, essentially trying to reverse engineer the process from the examples in my professor's slides and the groundwork rules laid out in the textbook. I hope I'm on the right track, and would greatly appreciate any insight you could help give me!

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    $\exists x \in \{a_0, a_1, a_2, \dots\}~F(x)$ is the same as $F(a_0) \lor F(a_1) \lor F(a_2) \lor \dots$. Similarly, $\forall x \in \{a_0, a_1, a_2, \dots\}~F(x)$ is the same as $F(a_0) \land F(a_1) \land F(a_2) \land \dots$. The whole thing is finite so just evaluate it. – 2017-01-14

1 Answers 1

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You get a lot right, but you also make two fundamental mistakes:

For the first example you can't assume that the existential claim has to be instantiated by $a$ ... It could of course also be $b$ or $c$, so it would be $\forall x Rxa \lor \forall x Rxb \lor \forall x Rxc$, and then that would work out to $(Raa \land Rba \land Rca) \lor (Rab \land Rbb \land Rcb)\lor (Rac \land Rbc \land Rcc)$. Now, this statement would still be false under this interpretation, but clearly it is quite different from what you got, since you only considered the case of $a$. For all the other examples you need to do something similar. In sum: for existentials you need to consider all possible objects in the domain just as much as you have to do for universals; it's just that the existentials become like 'or''s while universals become 'and's.

But this is still not the right way to think about this. What you need to do, is focus on the objects in the domain (i.e. the 1,2,3), rather than the constants in your language, because in general you may not have a constant for each object in your domain, and a quantifier makes a claim about all objects, not about all constant symbols. (In other words, you are kind of 'lucky' that in this case you do have a constant for each object, thus allowing you to rewrite the quantificational claims the way we just did). So to really do this right, you need to do this:

$\exists y \forall x Rxy$ is true under this interpretation if and only if there is some object in the domain such that all objects in the domain stand in the 'R' relation to it ... i.e. iff either the three tuples (1,1) , (2,1), and (3,1) are in R, or the hree tuples (1,2),(2,2), and (3,2) are in R, or the three tuple (1,3),(2,3), and (3,3) are in R ... which is of course still not the case, and hence the claim is false.

In other words: make sure to focus on the objects 1,2,3 in the domain, rather than the constants a,b,c, when you do semantical analysis!

In sum, I have some bad news for you: things are even more tedious than you already feared they were!