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Humanities Libertexts

7.3: Applying the Rules

Now let's apply our rules to some more involved examples. Let's try the argument

(Vx)Lxe v (Vx)~Lxa
~Lae                        
~(∃x)Lxa

I am going to write out the completed tree so that you can follow it as I explain each step. Don't try to understand the tree before I explain it. Skip over it, and start reading the explanation, referring back to the tree in following the explanation of each step.

7-3-1.png

   We begin by listing the premises and the negation of the conclusion. Our first move is to apply the rule for double negation to line 3, giving line 4. Next we work on line 1. Notice that even though '(Vx)' is the first symbol to appear on line 1, the sentence is not a universally quantified sentence. Ask yourself (As in chapters 8 and 9 in volume I): What is the last thing I do in building this sentence up from its parts? You take '(Vx)Lxe' and '(Vx)~Lxa' and form a disjunction out of them. So the main connective is a disjunction, and to make this sentence true in the interpretation we are building, we must apply the rule for disjunction, just as we used it in sentence logic. This gives line 5.

   In lines 1 through 4 our tree has one path. Line 5 splits this path into two branches. Each branch has its own universally quantified sentence which we must make true along the branch. Each branch also has '(∃x)Lxa', which is common to both branches and so must be made true along both branches. What should we do first?

   When we work on '(∃x)Lxa' we will have to introduce a new name. It is usually better to get out all the new names which we will have to introduce before working on universally quantified sentences. To see why, look at what would have happened if I had worked on line 5 before line 4. Looking at the right branch I would have instantiated '(Vx)~Lxa' with 'a' and 'e'. Then I would have returned to work on line 4, which would have introduced the new name 'c'. But now with a new name 'c' on the branch I must go back and instantiate (Vx)~Lxa' with 'c'. To make this sentence true, I must make it true for all instances. If a new name comes up in midstream, I must be sure to include its instance. Your work on a tree is more clearly organized if you don't have to return in this way to work on a universally quantified sentence a second time.

   We will see in the next chapter that in some problems we cannot avoid returning to work on a universally quantified sentence a second time. (It is because sometimes we cannot avoid this situation that we must never check a universally quantified sentence.) But in the present problem we keep things much better organized by following this practical guide:

Practical Guide: Whenever possible, work on existentially quantified sentences before universally quantified sentences.

   Now we can complete the problem. I work on line 4 before line 5. Line 4 is an existentially quantified sentence. The rule ∃ tells me to pick a new name, to use this new name in forming a substitution instance for the existentially quantified sentence, and to write this instance at the bottom of every open path on which the existentially quantified sentence appears. Accordingly, I pick 'c' as my new name and write the instance 'Lca' on each branch as line 6. Having done this, I check line 4, since I have now ensured that it will be made true along each open path on which it appears.

   Finally, I can work on line 5. On the left branch I must write substitution instances for '(Vx)Lxe' for all the names that appear along that branch. So below '(Vx)Lxe' I write 'Lae', 'Lee', and 'Lce', and I write the names 'a', 'e', and 'c' to the left of the target sentence '(Vx)Lxe' to note the fact that this sentence has been instantiated with these three names. The branch closes because 'Lae' of line 7 conflicts with '~Lae' on line 2. On the right branch I have '(Vx)~Lxa'. At the bottom of the branch I write its substitution instances for all names on the branch, giving '~Laa', '~Lea', and '-La'. Again, I write the names used to the left of the target sentence. '~Lca' is the negation of 'Lca' on line 6. So the right branch closes also, and the argument is valid.

   One more comment about this example: The new name requirement did not actually avoid any trouble in this particular case. If I had used either of the old names 'a' or 'e', I would in this case have gotten the right answer. Moreover, the tree would have been shorter. You may be thinking: What a bother this new name requirement is! Why should I waste my time with it in a case like this? But you must follow the new name requirement scrupulously if you want to be sure that the tree method works. When problems get more complicated, it is, for all practical purposes, impossible to tell whether you can safely get away without using it. The only way to be sure of always getting the right answer is to use the new name requirement every time you instantiate an existentially quantified sentence.

   Now let's try an example which results by slightly changing the first premises of the last case:

(Vx)(Lxe v ~Lxa)
~Lae                 
~(∃x)Lxa

Instead of starting with a disjunction of two universally quantified sentences, we start with a universal quantification of a disjunction:

7-3-2.png

Lines 1, 2, and 3 list the premises and the negation of the conclusion. Line 4 gives the result of applying ~~ to line 3. Looking at line 1, we ask, - What was the very last step performed in writing this sentence? The answer: applying a universal quantifier. So it is a universally quantified sentence. But line 4 is an existentially quantified sentence. Our practical guide tells us to work on the existentially before the universally quantified sentence. Accordingly, I pick a new name, 'c', and use it to instantiate '(∃x)Lxa', giving me line 5. Now I can return to line 1 and apply the rule V. At this point, the names on the branch are 'a', 'e', and 'c'. So I get the three instances of 1 written on lines 6, 7, and 8, and I record the names used to the left of line 1. Lines 9, 10, and 11 apply the v rules to lines 6, 7, and 8. Notice that I chose to work on line 8 before line 7. I am free to do this, and I chose to do it, because I noticed that the disjunction of line 8 would close on one branch, while the disjunction of line 7 would not close on any branches.

   We have applied the rules as far as they can be applied. No sentence can be made true by making shorter sentences true. We are left with two open branches, each of which represents a counterexample to the original argument. Let's write these counterexamples down.

   The branch labeled (i) at the bottom has the names 'e', 'c', and 'a'. (In principle, the order in which you list information on a branch makes no difference. But it's easiest to read off the information from the bottom of the branch up.) So I indicate the domain of branch (i)'s interpretation by writing D = {e,c,a}. What is true of e, c, and a in this interpretation? The branch tells us that e bears L to itself, that c bears L to e, that a does not bear L to itself, that c bears L to a and that a does not bear L to e. In short, the interpretation is

D = {e,c,a); Lee & Lce & ~Laa & Lca & ~Lae

   To read an interpretation of an open branch, you need only make a list of the branch's names and the atomic and negated atomic sentences which appear along the branch. We use the format I have just indicated to make clear that the names are names of the objects of the domain, and the atomic and negated atomic sentences describe what is true of these objects. Check your understanding by reading the counterexample off the branch labeled (ii). You should get

D = {e,a,c); ~Lea & Lce & ~Laa & Lca & ~Lae

   Notice that neither of these counterexamples as read off the branches constitutes complete interpretations. The branches fail to specify some of the atomic facts that can be expressed with 'L', 'a', 'c', and 'e'. For example, neither branch tells us whether 'LC' is true or false. We have seen the same situation in sentence logic when sometimes we had a relevant sentence letter and an open branch on which neither the sentence letter nor its negation appeared as the entire sentence at a point along the branch. Here, as in sentence logic, this happens when a branch succeeds in making everything along it true before completing a selection of truth values for all relevant atomic sentences. In effect, the branch represents a whole group of interpretations, one for each way of completing the specification of truth values for atomic sentences which the branch does not mention. But for our purposes it will be sufficient to read off the interpretation as the branch presents it and call it our counterexample even though it may not yet be a complete interpretation.

Exercise

7-3. 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) (∃x)(Px 3 Qx)             b) (∃x)Cx             c) (∃x)Jx v (∃x)Kx
    ~(Vx)(Px & ~Qx)          ~(∃x)~Cx                   (Vx)~Jx
                                                                       ~(Vx)~kx

 

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