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5.6: The Existential Elimination Rule

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    VI and ∃E are difficult rules. Many of you will have to work patiently - over this material a number of times before you understand them clearly. But if you have at least a fair understanding of VI, we can proceed to ∃E because ultimately these two rules need to be understood together.

    Let's go back to the first example in Section 5.4: Everyone likes either rock music or country-western. Someone does not like rock. So someone likes country-western. I will symbolize this as

    (Vx)(Rx v Cx)

    In informally showing this argument's validity, I used 'Doe', which I will now write just as 'd', as a stand-in name for the unknown "someone" who does not like rock. But I must be careful in at least two respects:

    1. I must not allow myself to apply VI to the stand-in name, 'd'. Otherwise, I could argue from '(∃x)~Rx' to '~Rd' to '(Vx)~Rx'. In short, I have to make sure that such a name never gets a hat.
    2. When I introduce the stand-in name, 'd', I must not be assuming anything else about the thing to which 'd' refers other than that '-R' is true of it.

    It's going to take a few paragraphs to explain how we will meet these two requirements. To help you follow these paragraphs, I'll begin by writing down our example's derivation, which you should not expect to understand until you have read the explanation. Refer back to this example as you read:

    (Vx)(Rx v Cx) 1 | (Vx)(Rx v Cx) P
    (∃x)~Rx 2 | (∃x)~Rx P
    (∃x)Cx 3 | d | ~Rd A
    4 | | (Vx)(Rx v Cx) 1, R
    5 | | Rd v Cd 4, VE
    6 | | Cd 3, 5, vE
    7 | | (∃x)Cx 6, ∃I
    8 | (∃x)Cx 2, 3-7, ∃E

    I propose to argue from the premise, '(∃x)~Rx', by using the stand-in name, 'd'. I will say about the thing named by 'd' what '(∃x)~Rx' says about "someone". But I must be sure that 'd' never gets a hat. How can I guarantee that? Well, names that occur in assumptions can't get hats anywhere in the subderivation governed by the assumption. So we can guarantee that 'd' won't get a hat by introducing it as an assumption of a subderivation and insisting that 'd' never occur outside that subderivation. This is what I did in line 3. '~Rd' appears as the subderivation's assumption, and the 'd' written just to the left of the scope line signals the requirement that 'd' be an Isolated Name. That is to say, 'd' is isolated in the subderivation the scope line of which is marked with the 'd'. An isolated name may never appear outside its subderivation.

    Introducing 'd' in the assumption of a subderivation might seem a little strange. I encounter the sentence, '(∃x)~Rx', on a derivation. I reason: Let's assume that this thing of which '~R' is true is called 'd', and let's record this assumption by starting a subderivation with '-Rd' as its assumption, and see what we can derive. Why could this seem strange? Because if I already know '(∃x)~Rx', no further assumption is involved in assuming that there is something of which '-R' is true. But, in a sense, I do make a new assumption in assuming that this thing is called 'd'. It turns out that this sense of making a special assumption is just what we need.

    By making 'd' occur in the assumption of a subderivation, and insisting that 'd' be isolated, that it appear only in the subderivation, I guarantee that 'd' never gets a hat. But this move also accomplishes our other requirement: If 'd' occurs only in the subderivation, 'd' cannot occur in any outer premise or assumption.

    Now let's see how the overall strategy works. Look at the argument's subderivation, steps 3-7. You see that, with the help of reiterated premise 1, from '~Rd' I have derived '(∃x)Cx'. But neither 1 nor the conclusion '(∃x)Cx' uses the name 'd'. Thus, in this subderivation, the fact that I used the name 'd' was immaterial. I could have used any other name not appearing in the outer derivation. The real force of the assumption '~Rd' is that there exists something of which '~R' is true (there is someone who does not like rock). But that there exists something of which '~R' is true has already been given to me in line 2! Since the real force of the assumption of line 3 is that there exists something of which '~R' is true, and since I am already given this fact in line 2, I don't really need the assumption 3. I can discharge it. In other words, if I am given the truth of lines 1 and 2, I know that the conclusion of the subderivation, 7, must also be true, and I can enter 7 as a further conclusion of the outer derivation.

    It is essential, however, that 'd' not appear in line 7. If 'd' appeared in the final conclusion of the subderivation, then I would not be allowed to discharge the assumption and enter this final conclusion in the outer derivation. For if 'd' appeared in the subderivation's final conclusion, I would be relying, not just on the assumption that '~R' was true of something, but on the assumption that this thing was named by 'd'.

    The example's pattern of reasoning works perfectly generally. Here is how we make it precise:

    A name is Isolated in a Subderivation if it does not occur outside the subderivation. We mark the isolation of a name by writing the name at the top left of the scope line of its subderivation. In applying this definition, remember that a sub-sub-derivation of a subderivation counts as part of the subderivation.

    Existential Elimination Rule: Suppose a sentence of the form (∃u)(. . . u. . .) appears in a derivation, as does a subderivation with assumption (. . . s . . .), a substitution instance of (∃u)(. . . u . . .). Also suppose that s is isolated in this subderivation. If X is any of the subderivation's conclusions in which s does not occur, you are licensed to draw X as a further conclusion in the outer derivation, anywhere below the sentence (∃u)(. . . u . . .) and below the subderivation. Expressed with a diagram:

    Where (. . . s . . .) is a substitution instance of (∃u) (. . . u . . .) and s is isolated in the derivation.

    | (∃u)(. . . u . . .)
    | s | (. . . s . . .)
    | | .
    | | .
    | | .
    | | x
    | X ∃E

    When you annotate your application of the ∃E rule, cite the line number of the existentially quantified sentence and the inclusive line numbers of the subderivation to which you appeal in applying the rule.

    You should be absolutely clear about three facets of this rule. I will illustrate all three.

    Suppose the ∃E rule has been applied, licensing the new conclusion, X, by appeal to a sentence of the form (∃u)(. . . u . . .) and a subderivation beginning with assumption (. . . s . . .):

    1) s cannot occur in any premise or prior assumption governing the
    2) s cannot occur in (∃u)(. . . u . . .), and
    3) s cannot occur in X.

    All three restrictions are automatically enforced by requiring s to be isolated in the subderivation. (Make sure you understand why this is correct.) Some texts formulate the ∃E rule by imposing these three requirements separately instead of requiring that s be isolated. If you reach chapter 15, you will learn that these three restrictions are really all the work that the isolation requirement needs to do. But, since it is always easy to pick a name which is unique to a subderivation, I think it is easier simply to require that s be isolated in the subderivation.

    Let us see how things go wrong if we violate the isolation requirement in any of these three ways. For the first, consider:

    Ca 1 | Ca P
    (∃x)Bx (Invalid!) 2 | (∃x)Bx P
    (∃x)(Cx &Bx) 3 | a| Ba A
    4 | | Ca 1, R
    5 | | Ca & Ba 3, 4, &I
    6 | | (∃x)(Ca & Bx) 5,∃I
    7 | (∃x)(Cx &Bx)

    Mistaken attempt to apply ∃E to 2 and 3-6. 'a' occurs in premise 1 and is not isolated in the sub derivation.

    From the fact that Adam is clever and someone (it may well not be Adam) is blond, it does not follow that any one is both clever and blond.
    Now let's see what happen if one violates the isolation requirement in the second way:

    (Vx)(∃y)Lxy 1 | (Vx)(∃y)Lxy P
    (∃x)Lxx (Invalid!) 2 | (∃y)Lây 1, VE
    3 | a| Laa A
    4 | | (∃x)Lxx 3, ∃I
    5 | (∃x)Lxx
    Mistaken attempt to apply ∃E to 2 and 3-4, 'a' occurs in 2 and is not isolated in the subderivation.

    From the fact that everyone loves someone, it certainly does not follow that someone loves themself.
    And, for violation of the isolation requirement in the third way:

    (∃y)Bx 1 | (∃x)Bx P
    (Vx)Bx (Invalid!) 2 | a| Ba A
    3 | | Ba 2, R
    4 | Bâ
    5 | (Vx)Bx
    Mistaken attempt to apply ∃E to 1 and 2-3, 'a' occurs in 4 and is not isolated in the subderivation.

    From the fact that someone is blond, it will never follow that everyone is blond.

    One more example will illustrate the point about a sub-sub-derivation being part of a subderivation. The following derivation is completely correct:

    1 | (Vx)(Cx ⊃ ~Bx) P
    2 | (∃x)Bx P
    3 | d |Bd A
    4 | | | Cd A
    5 | | | (Vx)(Cx ~Bx) 1, R
    6 | | | Cd ⊃~Bd 5, VE
    7 | | | ~Bd 4, 6, ⊃E
    8 | | 3| Bd 3, R
    9 | | ~Cd 4-8, ~I
    0 | 2| (∃x)~Cx 9, ∃I
    11 1| (∃x)~Cx 2, 3-10, ∃E

    You might worry about this derivation: If 'd' is supposed to be isolated in subderivation 2, how can it legitimately get into sub-sub-derivation 3?

    A subderivation is always part of the derivation in which it occurs, and the same holds between a sub-sub-derivation and the subderivation in which it occurs. We have already encountered this fact in noting that the premises and assumptions of a derivation or subderivation always apply to the derivation's subderivations, its sub-sub-derivations, and so on.

    Now apply this idea about parts to the occurrence of 'd' in sub-subderivation 3 above: When I say that a name is isolated in a subderivation I mean that the name can occur in the subderivation and all its parts, but the name cannot occur outside the subderivation.

    Here is another way to think about this issue: The 'd' at the scope line of the second derivation means that 'd' occurs to the right of the scope line and not to the left. But the scope line of subderivation 3 is not marked by any name. So the notation permits you to use 'd' to the right of this line also.

    I hope that you are now beginning to understand the rules for quantifiers. If your grasp still feels shaky, the best way to understand the rules better is to go back and forth between reading the explanations and practicing with the problems. As you do so, try to keep in mind why the rules are supposed to work. Struggle to see why the rules are truth preserving. By striving to understand the rules, as opposed to merely learning them as cookbook recipes, you will learn them better, and you will also have more fun.


    5-6. There is one or more mistakes in the following derivation. Write the hats where they belong, justify the steps that can be justified, and identify and explain the mistake, or mistakes.

    1 | (Vy)(∃y)Lxy P
    2 | (∃y)Lxb
    3 | | Lab A
    4 | | (Vx)Lay
    5 | | (∃x)(Vy)Lxy
    6 | (∃x)(Vy)Lxy

    5-7. Provide derivations which establish the validity of the following arguments:

    a) (∃x)Ix b) (∃x)(A ⊃ Px) c) (∃x)Hmx
    (Vx)(Ix ⊃ Jx) A ⊃ (∃x)Px (Vx)(~Hmx v Gxn)
    (∃x)Jx (∃x)Gxn

    d) (∃x)(Cfx &Cxf) e) (∃x)(Px vQx) f) (∃x)Px v (∃x)Qx
    (∃x)Cfx & (∃x)Cxf (∃x)Px v (∃x)Qx (∃x)(Px v Qx)

    g) (∃x)(Px ⊃ A) h) (V)(Px ⊃ A) i) (∃x)(Lxa ≡ Lex)
    (Vx)Px ⊃ A (∃x)Px ⊃ A (VVx)Lxa

    j) (Vx)(Gsx ⊃ ~Gxs)
    (∃x)Gxs ⊃ (∃x)~Gsx

    k) (∃x)(Px v Qx) l) (∃x)(~Mxt v Mtx) m) (∃x)Hxg v (∃x)Nxf
    (Vx)(Px ⊃ kx) (∃x)(Mtx ⊃ Axx) (Vx)(Hxg ⊃ Cx)
    (Vx)(Qx ⊃ Kx) (∃x)(~Mxt v Axx) (Vx)(Nxf ⊃ Cx)
    (∃x)kx (∃x)Cx

    n) (Vx)[(Fx v Gx) ⊃ Lxx] o) (Vx){Fx ⊃ (Rxa v Rax]
    (∃x)~(Lxx) (∃x)~Rxa
    (∃x)~Fx & (∃x)~Gx (Vx)~Rax ⊃ (∃x)~Fx

    p) (∃x)Qxj q) (Vx)~Fx r) (∃x)~Fx
    (∃x)(Qxj v Dgx) ⊃ (Vx)Dgx ~(∃x)Fx ~(Vx)Fx
    (Vx)(Dgx v Qjx)

    s) (Vx)(Jxx ⊃ ~Jxf) t) (∃x)Px v Qa u) A ⊃ (∃x)Px
    ~(∃x)(Jxx & Jxf) (Vx)~Px (∃x)(A ⊃ Px)

    5-8. Are you bothered by the fact that ∃E requires use of a subderivation with an instance of the existentially quantified sentence as its assumption? Good news! Here is an alternate version of ∃E which does not require starting a subderivation:

    | (∃u)(. . . u . . .)
    | (Vu)(. . . u . . .)
    | X ∃E

    Show that, in the presence of the other rules, this version is exchangeable with the ∃E rule given in the text. That is, show that the above is a derived rule if we start with the rules given in the text. And show that if we start with all the rules in the text except for ∃E, and if we use the above rule for ∃E, then the ∃E of the text is a derived rule.

    chapter summary Exercise

    Here is a list of important terms from this chapter. Explain them briefly and record your explanations in your notebook:

    a) Truth Preserving Rule of Inference
    b) Sound
    c) Complete
    d) Stand-in Name
    e) Govern
    f) Arbitrary Occurrence
    g) Existential Generalization
    h) Universal Generalization
    i) Isolated Name
    j) Existential Introduction Rule
    k) Existential Elimination Rule
    l) Universal Introduction Rule
    m) Universal Elimination Rule

    5.6: The Existential Elimination Rule is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

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