5.5: The Universal Introduction Rule
 Page ID
 1832
Here is the intuitive idea for universal introduction, as I used this rule in the soft music example: If a name, as it occurs in a sentence, is completely arbitrary, you can Universally Generalize on the name. This means that you rewrite the sentence with a variable written in for all occurrences of the arbitrary name, and you put a universal quantifier, written with the same variable, in front. To make this intuition exact, we have to say exactly when a name is arbitrary and what is involved in universal generalization. We must take special care because universal generalization differs importantly from existential generalization.
Let's tackle arbitrariness first. When does a name not occur arbitrarily? Certainly not if some assumption is made about (the object referred to by) the name. If some assumption is made using a name, then the name can't refer to absolutely anything. If a name occurs in a premise or assumption, the name can refer only to things which satisfy that premise or assumption. So a name does not occur arbitrarily when the name appears in a premise or an assumption, and it does not occur arbitrarily as long as such a premise or assumption is in effect.
The soft music example shows these facts at work. I'll use 'Rx' for 'x likes rock.', 'Cx' for 'x likes country/western.', and 'Sx' for 'x likes soft music.' Here are the formalized argument and derivation which I am going to use to explain these ideas:
(Vx)(Rx v Cx) 1  (Vx)(Rx v Cx) P
(Vx)(Cx ⊃ Sx) 2  (Vx)(Cx ⊃ Sx) P
(Vx)(~Rx ⊃ Sx) 3  Ra v Ca 1, VE
4  Ca ⊃ Sa 2, VE
5   ~Ra A
6   Ra v Ca 3, R
7   Ca 5, 6, vE
8   Ca ⊃ Sa 4, R
9   Sa 7, 8, ⊃E
10  ~Ra ⊃ Sa 59, ⊃I
11  (Vx)(~Rx ⊃ Sx) 10, VI
Where does 'a' occur arbitrarily in this example? It occurs arbitrarily in lines 3 and 4, because at these lines no premise or assumption using 'a' is in effect. We say that these lines are Not Governed by any premise or assumption in which 'a' occurs. In lines 5 through 9, however, 'a' does not occur arbitrarily. Line 5 is an assumption using 'a'. In lines 5 through 9, the assumption of line 5 is in effect, so these lines are governed by the assumption of line 5. (We are going to need to say that a premise or assumption always governs itself.) In all these lines something special is being assumed about the thing named by 'a', namely, that it has the property named by 'R'. So in these lines the thing named by 'a' is not just any old thing. However, in line 10 we discharge the assumption of line 5. So in line 10 'a' again occurs arbitrarily. Line 10 is only governed by the premises 1 and 2, in which 'a' does not occur. Line 10 is not governed by the assumption of line 5.
I am going to introduce a device to mark the arbitrary occurrences of a name. If a name occurs arbitrarily we will put a hat on it, so it looks like this: I. Marking all the arbitrary occurrences of 'a' in the last derivation ' makes the derivation look like this:
1  (Vx)(Rx v Cx) P
2  (Vx)(Cx ⊃ Sx) P
3  Râ v Câ 1, VE
4  Câ ⊃ Sâ 2, VE
5   ~Ra A
6   Ra v Ca 3, R
7   Ca 5, 6, vE
8   Ca ⊃ Sa 4, R
9   Sa 7, 8, ⊃E
10  ~Râ ⊃ Sâ 59, ⊃I
11  (Vx)(~Rx ⊃ Sx) 10, VI
Read through this copy of the derivation and make sure you understand why the hat occurs where it does and why it does not occur where it doesn't. If you have a question, reread the previous paragraph, remembering that a hat on a name just means that the name occurs arbitrarily at that place.
I want to be sure that you do not misunderstand what the hat means. A name with a hat on it is not a new kind of name. A name is a name is a name, and two occurrences of the same name, one with and one without a hat, are two occurrences of the same name. A hat on a name is a kind of flag to remind us that at that point the name is occurring arbitrarily. Whether or not a name occurs arbitrarily is not really a fact just about the name. It is a fact about the relation of the name to the derivation in which it occurs. If, at an occurrence of a name, the name is governed by a premise or assumption which uses the same name, the name does not occur there arbitrarily. It is not arbitrary there because the thing it refers to has to satisfy the premise or assumption. Only if a name is not governed by any premise or assumption using the same name is the name arbitrary, in which case we mark it by dressing it with a hat.
Before continuing, let's summarize the discussion of arbitrary occurrence with an exact statement:
Suppose that a sentence, X, occurs in a derivation or subderivation. That occurrence of X is Governed by a premise or assumption, Y, if and only if Y is a premise or assumption of X's derivation, or of any outer derivation of X's derivation (an outer derivation, or outerouter derivation, and so on). In particular, a premise or assumption is always governed by itself.
A name Occurs Arbitrarily in a sentence of a derivation if that occurrence of the sentence is not governed by any premise or assumption in which the name occurs. To help us remember, we mark an arbitrary occurrence of a name by writing it with a hat.
The idea for the universal introduction rule was that we would Universally Generalize on a name that occurs arbitrarily. We have discussed arbitrary occurrence. Now on to universal generalization.
The idea of a universal generalization differs in one important respect from the idea of an existential generalization. To see the difference, you must be clear about what we want out of a generalization: We want a new quantified sentence which follows from a sentence with a name.
For the existential quantifier, '(Ǝx)Lxx', '(Ǝx)Lax', and '(Ǝx)Lxa' all follow from 'Laa'. From the fact that Adam loves himself, it follows that Adam loves someone, someone loves Adam, and someone loves themself.
Now suppose that the name 'â' occurs arbitrarily in 'Lââ'. We know that "Adam" loves himself, where Adam now could be just anybody at all. What universal fact follows? Only that '(Vx)Lxx', that everyone loves themself. It does not follow that '(Vx)Lâx' or '(Vx)Lxâ'. That is, it does not follow that Adam loves everyone or everyone loves Adam. Even though 'Adam' occurs arbitrarily, '(Vx)Lâx' and '(Vx)Lxâ' make it sound as if someone ("Adam") loves everyone and as if someone ("Adam") is loved by everyone. These surely do not follow from 'Lââ'. But ƎI would license us to infer these sentences, respectively, from '(Vx)Lâx' and from '(Vx)Lxâ'.
Worse, â is still arbitrary in '(Vx)Lâx'. So if we could infer '(Vx)Lâx' from 'Lââ', we could then argue that in '(Vx)Lâx', 'â' could be anyone. We would then be able to infer '(Vy)(Vx)Lyx', that everyone loves everyone! But from 'Lââ' we should only be able to infer '(Vx)Lxx', that everyone loves themself, not '(Vy)(Vx)Lyx', that everyone loves everyone.
We want to use the idea of existential and universal generalizations to express valid rules of inference. The last example shows that, to achieve this goal, we have to be a little careful with sentences in which the same name occurs more than once. If s occurs more than once in (. . . s . . .), we may form an existential generalization by generalizing on any number of the occurrences of s. But, to avoid the problem I have just described and to get a valid rule of inference, we must insist that a universal generalization of (. . . s . . .), with respect to the name, s, must leave no instance of s in (. . . s . . .).
In other respects the idea of universal generalization works just like existential generalization. In particular, we must carefully avoid the trap of trying to replace a name by a variable already bound by a quantifier. This idea works exactly as before, so I will proceed immediately to an exact statement:
The sentence (Vu)(. . . u . . .) results by Universally Generalizing on the name s in (. . . s . . .) if and only if one obtains (Vu)(. . . u . . .) from (. . .s. ..)by
a) Deleting all occurrences of s in (. . . s . . .),
b) Replacing these occurrences with a variable, u, which is free at these occurrences, and
c) Applying (Vu) to the result.
(Vu)(. . . u . . .) is then said to be the Universal Generalization of (. . . s . . .) with Respect to the Name s.
With these definitions, we are at last ready for an exact statement of the universal introduction rule:
Universal Introduction Rule: If a sentence, X, appears in a derivation, and if at the place where it appears a name, ŝ, occurs arbitrarily in X, then you are licensed to conclude, anywhere below, the sentence which results by universally generalizing on the name ŝ in X. Expressed with a diagram:
Where j occurs arbitrarily in (. . . . . .) and (Vu)(. . ;u . . .) is the univers?l generalization . 1 of (. . . s . . .) with respect to s.
 (. . . ŝ . . .)
 (Vu)(. . . u . . .) VI
Where ŝ occurs arbitrarily in (. . . ŝ . . .) and (Vu)(. . . u. . .) is the universal generalization of (. . . ŝ . . .) with respect to ŝ.
Let's look at two simple examples to illustrate what can go wrong if you do not follow the rule correctly. The first example is the one we used to illustrate the difference between existential and universal generalization:
Everyone loves themself.
Everyone loves Adam. (Invalid!)
1  (Vx)Lxx P
2  Lââ 1,VE
3  (Vx)Lxâ
Mistaken attempt to apply VI to 2. 3 is not a universal generalization of 2
The second example will make sure you understand the requirement that VI applies only to an arbitrary occurrence of a name:
Adam is blond. 1  Ba P
Everyone is blond. (Invalid!) 2  (Vx)Bx
Mistaken attempt to apply VI to 1. 'a' is not arbitrary at 1.
The problem here is that the premise assumes something special about the thing referred to by 'a', that it has the property referred to by 'B'. We can universally generalize on a namethat is, apply VIonly when nothing special is assumed in this way, that is, when the name is arbitrary. You will see this even more clearly if you go back to our last formalization of the soft music example and see what sorts of crazy conclusions you could draw if you were to allow yourself to generalize on occurrences of names without hats.
Let's consolidate our understanding of VI by working through one more example. Before reading on, try your own hand at providing a derivation for
(Vx)(Lax & Lxa)
(Vx)(Lax ≡ Lxa)
If you don't see how to begin, use the same overall strategy we developed in chapter 6 of volume 1. Write a skeleton derivation with its premise and final conclusion and ask what you need in order to get the final, or target, conclusion.
1  (Vx)(Lax & Lxa) P
 ?
 ?
 (Vx)(Lax ≡ Lxa)
We could get our target conclusion by VI if we had a sentence of the form 'Lab̂ ≡ Lb̂a'. Let's write that in to see if we can make headway in this manner:
1  (Vx)(Lax & Lxa) P
 ?
 ?
 Lab̂ ≡Lb̂a
 (Vx)(Lax ≡ Lxa) VI
'Lab̂ ≡ Lb̂a' is now our target conclusion. As a biconditional, our best bet is to get it by ≡I from 'Lab̂ ⊃ Lb̂a' and 'Lb̂a ⊃ Lab̂'. (I didn't write hats on any names because, as I haven't written the sentences as part oE the derivation, I am not yet sure which sentences will govern these two conditionals.) The conditionals, in turn, I hope to get from two subderivations, one each starting from one of the antecedents of the two conditionals:
1  (Vx)(Lax & Lxa) P
 ?
  Lab A
  ?
  Lba
 Lab̂ ⊃ Lb̂a ⊃I
 ?
  Lba A
  ?
  Lab

 Lb̂a ⊃ Lab̂ ⊃I
 Lab ≡ Lba
 (Vx)(Lax ≡ Lxa) VI
Notice that 'b' gets a hat wherever it appears in the main derivation. There, 'b' is not governed by any assumption in which 'b' occurs. But 'b' occurs in the assumptions of both subderivations. So in the subderivations 'b' gets no hat. Finally, 'a' occurs in the original premise. That by itself rules out putting a hat on 'a' anywhere in the whole derivation, which includes all of its subderivations.
Back to the question of how we will fill in the subderivations. We need to derive 'Lba' in the first and 'Lab' in the second. Notice that if we apply VE to the premise, using 'b' to instantiate 'x', we get a conjunction with exactly the two new target sentences as conjuncts. We will be able to apply &E to the conjunction and then simply reiterate the conjuncts in the subderivations. Our completed derivation will look like this:
1  (Vx)(Lax & Lxa) P
2  Lab̂ & Lb̂a 1, VE
3  Lab̂ 2, &E
4  Lb̂s 2, &E
5   Lab A
6   Lba 4, R
7  Lab̂ ⊃ Lb̂a 56. ⊃I
8   Lba A
9   Lab 3, R
10  Lb̂a ⊃ Lab̂ 89, ⊃I
11  Lab ≡ Lba 7, 10, ≡I
12  (Vx)(Lax ≡ Lxa) 11, VI
Once more, notice that 'b' gets a hat in lines 2, 3, and 4. In these lines no premise or assumption using 'b' is operative. But in lines 5, 6, 8, and 9, 'b' gets no hat, even though exactly the same sentences appeared earlier (lines 3 and 4) with hats on 'b'. This is because when we move into the subderivations an assumption goes into effect which says something special about 'b'. So in the subderivations, off comes the hat. As soon as this special assumption about 'b' is discharged, and we move back out of the subderivation, no special assumption using 'b' is in effect, and the hat goes back on 'b'.
You may well wonder why I bother with the hats in lines like 2, 3, 4, 7, and 10, on which I am never going to universally generalize. The point is that, so far as the rules go, I am permitted to universally generalize on 'b' in these lines. In this problem I don't bother, because applying VI to these lines will not help me get my target conclusion. But you need to develop awareness of just when the formal statement of the VI rule allows you to apply it. Hence you need to learn to mark those places at which the rule legitimately could apply.
Students often have two more questions about hats. First, VI permits you to universally generalize on a name with a hat. But you can also apply ∃1 to a name with a hat. Now that I have introduced the hats, the last example in section 53 should really look like this:
1  (Vx)(Lxx P
2  Lââ 1, VE
3  (∃I)Lxx 2, ∃I
If everyone loves themself, then Arb loves him or herself, whoever Arb may be. But then someone loves themself. When a name occurs arbitrarily, the name can refer to anything. But then it also refers to something. You can apply either VI or ∃1 to a hatted name.
It is also easy to be puzzled by the fact that a name which is introduced in the assumption of a subderivation, and thus does not occur arbitrarily there, can occur arbitrarily after the assumption of the subderivation has been discharged. Consider this example:
1  (Vx)Px ⊃ (Vx)Qx P
2   Lab A
3   (∃I)Px 2, ∃I
4   (∃I)Px ⊃ (Vx)Qx 1, R
5   (Vx)Qx 3, 4, ⊃E
6   Qa 5, VE
7  Pâ ⊃ Qâ 26, ⊃I
8  (Vx)(Px ⊃ Qx) 7, VI
In the subderivation something is assumed about 'a', namely, that it has the property P. So, from the point of view of the subderivation, 'a' is not arbitrary. As long as the assumption of the subderivation is in effect, 'a' cannot refer to just anything. It can only refer to something which is P. But after the subderivation's assumption has been discharged, 'a' is arbitrary. Why? The rules tell us that 'a' is arbitrary in line 7 because line 7 is not governed by any premises or assumptions in which 'a' occurs. But to make this more intuitive, notice that I could have just as well constructed th~ sam: subderivation using the name 'b' instead of 'a', using ⊃E to write 'Pb̂ ⊃ Qb̂' on line 7. Or I could have used 'c', 'd', or any other name. This is why 'a' is arbitrary in line 7. I could have arrived at a conditional in line 7 using any name I liked instead of using 'a'.
Some students get annoyed and frustrated by having to learn when to put a hat on a name and when to leave it off. But it's worth the effort to learn. Once you master the hat trick, VI is simple: You can apply VI whenever you have a name with a hat. Not otherwise.
Exercise
54. There is a mistake in the following derivation. Put on hats where they belong, and write in the justification for those steps which are justified. Identify and explain the mistake.
1  (Vx)(Bx ⊃ Cx) P
2  Be ⊃ Ce
3   Be A
4   Be ⊃ Ce
5   Ce
6   (Vx)Ce
7  Be ⊃(Vx)Cx
55. Provide derivations which establish the validity of the following arguments. Be sure you don't mix up sentences which are a quantification of a sentence formed with a '&', a 'v', or a '⊃' with compounds formed with a '&', a 'v', or a '⊃', the components of which are quantified sentences. For example, '(Vx)(Px & Qa)' is a universally quantified sentence to which you may apply VE.'(Vx)Px & Qa' is a conjunction to which you may apply &E but not VE.
a) (Vx)(Fx & Gx) b) (Vx)(Mx ⊃ Nx) c) A
(Vx)Fx (Vx)Mx (Vx)(A v Nx)
(Vx)Nx
d) (Vx)Hx & (Vx)Qx e) (Vx)(Kxm & Kmx) f) (Vx)(Fx v Gx)
(Vx)(Hx & Qx) (Vx)Kxm & (Vx)Kmx (Vx)(Fx ⊃ Gx)
(Vx)Gx
g) (Vx)~Px v C h) (Vx)(Rxb ⊃ Rax) i) (Vx)(Gxh ⊃ Gxm)
(Vx)(~Px v C) (Vx)Rxb ⊃ (Vx)Rax (Vx)(~Gxm ⊃ Gxh)
j) (Vx)(Mx ⊃ Nx) k) T ⊃ (Vx)Mdx l) (Vx)(Hff ⊃ Lxx)
(Vx)(Nx ⊃ Ox) (Vx)(T ⊃ Mdx) Hff ⊃ (Vx)Lxx
(Vx)(Mx ⊃ Ox)
m) (Vx)Px v (Vx)Qx n) (Vx)Hx o) (Vx)(Sx ≡ Ox)
(Vx)(Px v Qx) (3x)Hx ⊃ (Vx)(Hx ⊃ Jx) (Vx)Sx ≡ (Vx)Ox
(Vx)x
p) (∃x)Px ⊃ A q) ~(∃x)Px r) ~(Vx)Px s) (Vx)Px ⊃ A
(Vx)(Px ⊃ A) (Vx)~Px (∃x)~Px (∃x)(Px ⊃ A)
t) ~(Vx)(Jx ⊃ ~Kx) u) ~(∃x)Qx v H v) ~(∃x)Dx
(∃x)(Jx & Kx) (Vx)(~Qx v H) (Vx)(Dx ⊃ Kx)