# 5.2: The Universal Elimination Rule

Consider the argument

Everyone is blond.      (Vx)Bx

Intuitively, if everyone is blond, this must include Adam. So if the premise is true, the conclusion is going to have to be true also. In terms of interpretations, let's consider any interpretation you like which is an interpretation of the argument's sentences and in which the premise, '(Vx)Bx', is true. The definition of truth of a universally quantified sentence tells us that '(Vx)Bx' is true in an interpretation just in case all of its substitution instances are true in the interpretation. Observe that 'Ba' is a substitution instance of '(Vx)Bx'. So in our arbitrarily chosen interpretation in which '(Vx)Bx' is true, 'Ba' will be true also. Since 'Ba' is true in any interpretation in which '(Vx)Bx' is true, the argument is valid.

(In this and succeeding chapters I am going to pass over the distinction between someone and something, as this complication is irrelevant to the material we now need to learn. I could give examples of things instead of people, but that makes learning very dull.)

The reasoning works perfectly generally:

Universal Elimination Rub: If X is a universally quantified sentence, then you are licensed to conclude any of its substitution instances below it. Expressed with a diagram, for any name, s, and any variable, u,

| (Vu)(. . . u . . .)
| .
| .
| .
| (. . . s . . .)            VE

Remember what the box and the circle mean: If on a derivation you encounter something with the form of what you find in the box, the rule licenses you to conclude something of the form of what you find in the circle.

Here is another example:

Everyone loves Eve.     (Vx)Lxe        1 | (Vx)Lxe      P
Adam loves Eve.          Lae              2 | Lae           1, VE

In forming the substitution instance of a universally quantified sentence, you must be careful always to put the same name everywhere for the substituted variable. Substituting 'a' for 'x' in '(Vx)Lxx', we get 'Laa', not 'Lxa'. Also, be sure that you substitute your name only for the occurrences of the variable which are free after deleting the initial quantifier. Using the name 'a' again, the substitution instance of '(Vx)(Bx  ⊃ (Vx)Lxe)' is 'Ba ⊃ (Vx)Lxe'. The occurrence of 'x' in 'Lxe' is bound by the second '(Vx)', and so is still bound after we drop the first '(Vx)'. If you don't understand this example, you need to review bound and free variables and substitution instances, discussed in chapter 3.

When you feel confident that you understand the last example, look at one more:
(Vx)(Gx ⊃ Kx)                1 | (Vx)(Gx ⊃ Kx)     P
Gf                                 2 | Gf                     P
Kf                                 3 | Gf ⊃Kf               1, VE
4 | Kf                     2, 3, ⊃E

Exercise

5-1. Provide derivations which demonstrate the validity of these arguments. Remember to work from the conclusion backward, seeing what you will need to get your final conclusions, as well as from the premises forward. In problem (d) be sure you recognize that the premise is a universal quantification of a conditional, while the conclusion is the very different conditional with a universally quantified antecedent.

a) (Vx)(Px & Dx)              b) (Vx)(Px & Dx)            c) (Vx)(Dx  Kx)
Pk                            Pd & Dk                         (Vx)Dx
ka

d) (Vx)(Mx ⊃ A)               e) (Vx)(Fx v Hx)            f) (Vx)(~Bx v Lcx)
(Vx)Mx ⊃ A                     (Vx)(Fx ⊃ Dx)                 (Vx)Bx ⊃ Lcd
(Vx)(Hx ⊃ Dx)
Dp & Db

g) (Vx)(Lxx ⊃ Lxh)             h) (Vx)(Rxx v Rxk)
~Lmh                             (Vy)~Ryk
~(Vx)Lxx                           Rcc & Rff