7.2: The Rule for Existential Quantification

Consider the argument

Somebody is blond. (∃x)Bx

As we noted in chapter 2, this argument is obviously invalid. If somebody is blond, it does not follow that Adam is blond. The blond might well be somebody else. We will have to keep the clear invalidity of this argument in mind while formulating the rule for existentially quantified sentences to make sure we get the rule right.
Begin by listing the premise and the negation of the conclusion:

1 (∃x)Bx P
2 ~Ba ~C

As in the last example, we already know that we have an interpretation started, this time with one object named 'a'. We also know that '~Ba' will have to be true in this interpretation. Can we extend the interpretation so as also to make '(∃x)Bx' true?

We have to be very careful here. We may be tempted to think along the following lines: An existentially quantified sentence is true in an interpretation just in case at least one of its substitution instances is true in the interpretation. We have one name, 'a', in the interpretation, so we could make '(3x)Bx' true by making its substitution instance, 'Ba', true. But if we do that, we add 'Ba' to a branch which already has '-Ba' on it, so that the branch would close. This would tell us that there are no counterexamples, indicating that the inference is valid. But we know the inference is invalid. Something has gone wrong.

As I pointed out in introducing the example, the key to the problem is that the blond might well be someone other than Adam. How do we reflect this fact in our rules? Remember that in extending a branch downward we are building an interpretation. In so doing, 'we are always free to add new objects to the interpretation's domain, which we do by bringing in new names in sentences on the tree. Since there is a possibility that the blond might be somebody else, we indicate this by instantiating our existentially quantified sentence with a new name. That is, we make '(∃x)Bx' true by writing 'Bb' at the bottom of the branch, with 'b' a new name. We bring the new name, 'b', into the interpretation to make sure that there is no conflict with things that are true of the objects which were in the interpretation beforehand.

The completed tree looks like this:

1 (∃x)Bx P
2 ~Ba ~C
3 Bb 1, ∃, New name

New name Invalid. Counterexample: D = {a,b); ~Ba & Bb

The open branch represents a counterexample. The counterexample is an interpretation with domain D = {a,b), formed with the names which appear on the open branch. The open branch tells us what is true about a and b in this interpretation, namely, that ~Bb & Bb.

You may be a little annoyed that I keep stressing 'new name'. I do this because the new name requirement is a very important aspect of the rule for existentially quantified sentences-an aspect which students have a very hard time remembering. When I don't make such a big fuss about it, at least 50 percent of a class forgets to use a new name on the next test. By making this fuss I can sometimes get the percentage down to 25 percent.

Here is the reason for the new name requirement. Suppose we are working on a sentence of the form (∃u)(. . . u . . .) such as '(∃x)Bx' in our example. And suppose we try to make it true along each open branch on which it appears by writing a substitution instance, (. . . t . . .), at the bottom of each of these branches. Now imagine, as happened in our example, that ~(. . . t . . .)-or something which logically implies ~(. . . t . . .)-already appears along one of these branches. In the example we already had '~Ba'. This would lead to the branch closing when in fact we can make a consistent interpretation out of the branch. We can always do this by instantiating (∃u)(. . . u . . .) with a new name, say, s, a name which does not appear anywhere along the branch. We use this new name in the instantiation (. . . s . . .). Then (. . . s . . .) can't conflict with a sentence already on the branch, and we are guaranteed not to have the kind of trouble we have been considering.

Not infrequently you get the right answer to a problem even if you don't use a new name when instantiating an existentially quantified sentence. But this is just luck, or perhaps insight into the particular problem, but insight which cannot be guaranteed to work with every problem. We want the rules to be guaranteed to find a counterexample if there is one. The only way to guarantee this is to write the rule for existentially quantified sentences with the new name requirement. This guarantees that we will not get into the kind of difficulty which we illustrated with our example:

Rule ∃: If an existentially quantified sentence (∃u)(. . . u. . .) appears as the entire sentence at some point on a tree, do the following to each open branch on which (∃u)(. . . u . . .) appears: First pick a new name, s, that is, a name which does not appear anywhere on the branch. Then write the one substitution instance (. . . s . . .) at the bottom of the branch. Put a check by (∃u)(. . . u . . .).

Why do we always need a new name for an' existentially quantified sentence but no new name for a universally quantified sentence (unless there happens to be no names)? In making a universally quantified sentence true, we must make it true for all things (all substitution instances) in the interpretation we are building. To make it true for more things, to add to the interpretation, does no harm. But it also does no good. If a conflict is going to come up with what is already true in the interpretation, we cannot avoid the conflict by bringing in new objects. This is because the universally quantified sentence has to be true for all the objects in the interpretation anyway.

We have an entirely different situation with existentially quantified sentences. They don't have to be true for all things in the interpretation. So they present the possibility of avoiding conflict with what is already true in the interpretation by extending the interpretation, by making each existentially quantified sentence true of something new. Finally, since the rules have the job of finding a consistent interpretation if there is one, the rule for existentially quantified sentences must incorporate this conflict-avoiding device.

Exercise

7-2. Test the following arguments for validity. State whether each argument is valid or invalid; when invalid, give the counterexamples shown by the open paths.

a) A ⊃ (∃x)Gx b) (∃x)Dx c) (∃x)(Px & Qx)
A (∃x)~Dx Pa v Qb
Gb A

d) (∃x)Px e) A v B
(∃x)Qx A ⊃ (∃x)Nx
Pm ≡ Qm B ⊃ (∃x)Nx
Ng