Skip to main content
Humanities Libertexts

5.3: The Existential Introduction Rule

Consider the argument

        Adam is blond.             Ba      
        Someone is blond.        (Ǝx)Bx

Intuitively, this argument is valid. If Adam is blond, there is no help for it: Someone is blond. Thinking in terms of interpretations, we see that this argument is valid according to our new way of making the idea of validity precise. Remember how we defined the truth of an existentially quantified sentence in an interpretation: '(Ǝx)Bx' is true in an interpretation if and only if at least one of its substitution instances is true in the interpretation. But 'Ba' is a substitution instance of '(Ǝx)Bx'. So, in any interpretation in which 'Ba' is true, '(Ǝx)Bx' is true also, which is just what we mean by saying that the argument "Ba. Therefore (Ǝx)Bx." is valid.

   You can probably see the form of reasoning which is at play here: From a sentence with a name we can infer what we will call an Existential Generalization of that sentence. '(Ǝx)Bx' is an existential generalization of 'Ba'. We do have to be a little careful in making this notion precise because we can get tripped up again by problems with free and bound variables. What would you say is a correct existential generalization of '(Vx)Lax'? In English: If Adam loves everyone, then we know that someone loves everyone. But we have to use two different variables to transcribe 'Someone loves everyone': '(Ǝy)(Vx)Lyx'. If I start with '(Vx)Lax', and replace the 'a' with 'x', my new occurrence of 'x' is bound by that universal quantifier. I will have failed to generalize existentially on 'a'.

   Here is another example for you to try: Existentially generalize

(i) Ba ⊃ (Vx)  Lax
     2       3     45

If I drop the 'a' at 2 and 4, write in 'x', and preface the whole with '(Ǝx)', I get

(ii) (Ǝx)(Bx ⊃ (Vx)Lxx)     Wrong
       1    2        3   45

The 'x' at 4, which replaced one of the 'a's, is bound by the universally quantified 'x' at 3, not by the existentially quantified 'x' at 1, as we intend in forming an existential generalization. We have to use a new variable. A correct existential generalization of 'Ba ⊃ (Vx)Lax' is

(iii) (Ǝy)(By ⊃ (Vx) Lyx)
        1    2       3     45

as are

(iv) (Ǝy)(By ⊃ (Vx) Lax)
        1    2       3     45

and

(iii) (Ǝy)(Ba⊃(Vx) Lyx)
        1    2     3     45

   Here is how you should think about this problem: Starting with a closed sentence, (. . . s . . .), which uses a name, s, take out one or more of the occurrences of the name s. For example, take out the 'a' at 4 in (i). Then look to see if the vacated spot is already in the scope of one (or more) quantifiers. In (i) to (v), the place marked by 4 is in the scope of the '(Vx)' at 3. So you can't use 'x'. You must perform your existential generalization with some variable which is not already bound at the places at which you replace the name. After taking out one or more occurrences of the name, s, in (. . . s . . . ), replace the vacated spots with a variable (the same variable at each spot) which is not bound by some quantifier already in the sentence.

   Continuing our example, at this point you will have turned (i) into

(vi) Ba ⊃ (Vx)Lya

You will have something of the form (. . . u . . .) in which u is free: 'y' is free in (vi). At this point you must have an open sentence. Now, at last, you can apply your existential quantifier to the resulting open sentence to get the closed sentence (Ǝu)(. . . u . . .).

   To summarize more compactly:

(Ǝu)(. . . u . . .) is an Existential Generalization of (. . . s . . .) with respect to the name s if and only if (Ǝu)(. . . u . . .) results from (. . . s . . .) by

a) Deleting any number of occurrences of s in (. . . s . . .),
b) Replacing these occurrences with a variable, u, which is free at these occurrences, and
c) Applying (Ǝu) to the result.

(In practice you should read (a) in this definition as "Deleting one or more occurrences of s in (. . . s . . .)." I have expressed (a) with "any number of' so that it will correctly treat the odd case of vacuous quantifiers, which in practice you will not need to worry about. But if you are interested, you can figure out what is going on by studying exercise 3-3.)

   It has taken quite a few words to set this matter straight, but once you see the point you will no longer need the words.

   With the idea of an existential generalization, we can accurately state the rule for existential introduction:

Existential Introduction Rule: From any sentence, X, you are licensed to conclude any existential generalization of X anywhere below. Expressed with a diagram,

| (. . . s . . .)   
| .
| .
| .
| (Ǝu)(. . . u . . .) ƎI

When (Ǝu)(. . . u . . .) is an existential generalization of (. . . s . . .)

   Let's look at a new example, complicated only by the feature that it involves a second name which occurs in both the premise and the conclusion:

Adam loves Eve.              Lae          
Adam loves someone.      (Ǝx)Lax

'(Ǝx)Lax' is an existential generalizaton of 'Lae'. So ƎI applies to make the following a correct derivation:

1 | Lae              P
2 | (Ǝx)Lax        1, ƎI

   To make sure you have the hang of rule ƎI, we'll do one more example. Notice that in this example, the second premise has an atomic sentence letter as its consequent. Remember that predicate logic is perfectly free to use atomic sentence letters as components in building up sentences.

ka                     1 | ka                        P
(Ǝx)Kx ⊃ P         2 | (Ǝx)Kx ⊃ P            P
P                       3 | (Ǝx)Kx                 1, ƎI
                         4 | P                         2, 3, ⊃E

In line 4 I applied 3E to lines 2 and 3. ⊃E applies here in exactly the same way as it did in sentence logic. In particular ⊃E and the other sentence logic rules apply to sentences the components of which may be quantified sentences as well as sentence logic sentences.

   Now let's try an example which applies both our new rules:

(Vx)Lxx               1 | (Vx)Lxx        P
(Ǝx)Lxx               2 | Laa              1, VE
                          3(Ǝx)Lxx        2, ƎI

   In addition to illustrating both new rules working together, this example illustrates something else we have not yet seen. In past examples, when I applied VE I instantiated a universally quantified sentence with a name which already occurred somewhere in the argument. In this case no name occurs in the argument. But if a universally quantified sentence is true in an interpretation, all of its substitution instances must be true in - the interpretation. And every interpretation must have at least one object in it. So a universally quantified sentence must always have at least one substitution instance true in an interpretation. Since a universally quantified sentence always has at least one substitution instance, I can introduce a name into the situation with which to write that substitution instance, if no name already occurs.

   To put the point another way, because every interpretation always has at least one object in it, I can always introduce a name to refer to some object in an interpretation and then use this name to form my substitution instance of the universally quantified sentence.

   Good. Let's try yet another example:

(Vx)(Cx ⊃ Mx)          1 | (Vx)(Cx  Mx)      P
Cd                           2 | Cd                    P
(Ǝx)Mx                     3 | Cd ⊃ Md           1, VE
                               4 | Md                   2, 3, ⊃E
                               5 | (Ǝx)Mx              4, ƎI

Notice that although the rules permit me to apply Ǝ1 to line 2, doing so would not have gotten me anywhere. To see how I came up with this derivation, look at the final conclusion. You know that it is an existentially quantified sentence, and you know that Ǝ1 permits you to derive such a sentence from an instance, such as 'Md'. So you must ask yourself: Can I derive such an instance from the premises? Yes, because the first premise says about everything that if it is C, then it is M. And the second premise says that d, in particular, is C. So applying VE to 1 you can get 3, which, together with 2, gives 4 by ⊃E.

Exercise 

5-2. Provide derivations which demonstrate the validity of the following arguments:

a)         Na                    b)  (Vx)(Kx & Px)             c) (Vx)(Hx ⊃ ~Dx)
      (Ǝx)(Nx v Gx)             (Ǝx)Kx & (Ǝx) Px                     Dg           
                                                                               (Ǝx)~Hx

d) (Vx)Ax & (Vx)Txd       e)    Fa v Nh                    f)  (Vx) Sx V Jx)  
     (Ǝx)(Ax & Txd                 (Ǝx)Fx v (Ǝx)Nx            (Ǝx) Sx v (Ǝx)Jx

g) (Ǝx)Rxa ⊃ (Vx)Rax       h)  Lae v Lea                  i) (Ǝx)Jx ⊃ Q
             Rea                       (Ǝx)Lax ⊃ A                       (Vx)Jx    
         (Ǝx)Rax                     (Ǝx)Lxa ⊃ A                           Q
                                                 A

j) (Vx)(Max v Mex)         k) (Vx)(Kxx ≡ Px)              l) (Vx)(~Oxx v Ix)
    ~(Ǝx)Max v Bg              (Vx){Kjx  & (Px ⊃Sx)}       (Vx)(Ix ⊃ Rxm)      
    ~(Ǝx)Mex v Bg                       (Ǝx)Sx                    (Ǝx)Oxx ⊃ (Ǝx)Rxm
         (Ǝx)Bx

  • Was this article helpful?