Skip to main content
Humanities LibreTexts

3.7.4: Negated Quantified Sentences

  • Page ID
    1845
  • \( \newcommand{\vecs}[1]{\overset { \scriptstyle \rightharpoonup} {\mathbf{#1}} } \) \( \newcommand{\vecd}[1]{\overset{-\!-\!\rightharpoonup}{\vphantom{a}\smash {#1}}} \)\(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\) \(\newcommand{\id}{\mathrm{id}}\) \( \newcommand{\Span}{\mathrm{span}}\) \( \newcommand{\kernel}{\mathrm{null}\,}\) \( \newcommand{\range}{\mathrm{range}\,}\) \( \newcommand{\RealPart}{\mathrm{Re}}\) \( \newcommand{\ImaginaryPart}{\mathrm{Im}}\) \( \newcommand{\Argument}{\mathrm{Arg}}\) \( \newcommand{\norm}[1]{\| #1 \|}\) \( \newcommand{\inner}[2]{\langle #1, #2 \rangle}\) \( \newcommand{\Span}{\mathrm{span}}\)\(\newcommand{\AA}{\unicode[.8,0]{x212B}}\)

    7-4. NEGATED QUANTIFIED SENTENCES

    To complete the rules for quantification, we still need the rules for the negation of quantified sentences. As always, we illustrate with a simple example:

    Lae I Lae P (3x)Lxe 2 -(3x)Lxe -C

    What will make line 2 true? All we need do is to make use of the equivalence rule for a negated existential quantifier, which we proved in section 34. (Remember: "Not one is" comes to the same thing as "All are not." If you have forgotten that material, reread the first few paragraphs of section 34you will very quickly remember how it works.) '-(3x)Lxe' is true in an interpretation if and only -if '(Vx)-Lxe' is true in the interpretation. So we can make '-(3x)Lxe' true by making '(Vx)-Lxe' true. This strategy obviously will work for any negated existentially quantified sentence. So we reformulate the -3 rule for logical equivalence as a rule for working on a negated existentially quantified sentence on a tree:

    Rule -3: If a sentence of the form -(3u)(. . . u . . .) appears as the full sentence at any point on a tree, write (Vu)-(. . . u . . .)at the bottom of every open branch on which -(3u)(. . . u . . .) appears. Put a check by -(3u)(. . . u . . .).

    With this rule, we complete our problem as follows:

    1 Lae P 42 -(3x)Lxe -C a, e, 3 (Vx)-Lxe 2, -3 4 -Lae 3, v 5 -Lee 3, v X Valid

    Note that I did not use a new name when I worked on line 2. The new name rule does not enter anywhere in this problem because the problem has no existentially quantified sentence as the full sentence at some point on the tree. Consequently, the rule 3 never applies to any sentence in this problem. Line 2 is the negation of an existentially quantified sentence, and we deal with such a sentence with our new rule, -3. Line 2 gives line 3 to which the rule V applies.

    The story for the negation of a universally quantified sentence goes the same way as for a negation of an existentially quantified sentence. Just as "not one is" comes to the same thing as "all are not," "not all are" comes to the same thing as "some are not." In other words, we appeal to the rule -V for logical equivalence in exactly the same way as we appealed to the rule -3 for logical equivalence. Reformulating for application to a negated universally quantified sentence on a tree, we have

    Rule -V: If a sentence of the form -(Vu)(. . . u . . .) appears as the full sentence at any point on a wee, write (31)-(. . . u . . .) at the bottom of

    140 Truth Trees for Predicate Logic: Fundamentab 7-4. Negated Quantijkd Sentences 121

    every open branch on which -(Vu)(. . . u . . .) appears. Put a check by -(Vu)(. . . u . . .).

    Once again, this rule works because -(Vu)(. . . u . . .) is equivalent to (3u)-(. . . u . . .), as we noted in section 3-4 and as you proved in exercise 3-5.

    Here is an example to illustrate the -V rule:

    (3x)Lxe J1 (3x)Lxe P J2 --(Vx)Lxe -C J3 (3x)-Lxe 2, -V 4 Lce 1, 3, New name 5 -Lde 3, 3, New name

    Invalid. Counterexample: D = {d,e,c); -Lde & Lce

    In this example, note that failure to follow the new name rule at step 5 would have incorrectly closed the branch. Also note that we do not instantiate line 2 with any of the names. Line 2 is not a universally quantified sentence. Rather, it is the negation of a universally quantified sentence which we treat with the new rule -V.

    Now you have all the rules for quantifiers. It's time to practice them.

    Before you go to work, let me remind you of the three most common mistakes students make when working on trees. First of all, you must be sure you are applying the right rule to the sentence you are working on. The key is to determine the sentence's main connective. You then apply the rule for that connective (or for the first and second connectives in case of negated sentences). You should be especially careful with sentences which begin with a quantifier. Some are quantified sentences, some are not; it depends on the parentheses. '(Vx)(Px > A)' is a universally quantified sentence, so the rule V applies to it. It is a universally quantified sentence because the initial universal quantifier applies to the whole following sentence as indicated by the parentheses. By way of contrast, '(Vx)(Lxa 3 Ba) & Lba' is not a universally quantified sentence. It is a conjunction, and the rule & is the rule to apply to it.

    The second mistake to watch for especially carefully is failure to instantiate a universally quantified sentence with all the names that appear on the sentence's branch. When a universally quantified sentence appears on a branch, you are not finally done with the sentence until either the branch closes or you have instantiated it with all the names that appear on the branch. - - 1

    Finally, please don't forget the new name requirement when you 1 work on an existentially quantified sentence. When instantiating an existentially quantified sentence,. you use only one name, but that I name must not yet appear anywhere on the branch. 1

    7-4. Use the truth tree method to test the following arguments for validity. In each problem, state whether or not the argument is valid; if invalid give a counterexample. al) (Vx)Px & (Vx)Qx a2) (Vx)(Px & Qx) (Vx)(Px & Qx) (Vx)Px & (Vx)Qx bl ) (3x)Px v (3x)Qx b2) (3x)(Px v Qx) (~X)(PX v Qx) (3x)Px v (3x)Qx el) A > (Vx)Px e2) (Vx)(A > Px) (Vx)(A > Px) A 3 (Vx)Px fl ) A 3 (3x)Px f2) (3x)(A > Px) (3x)(A > Px) A 3 (3x)Px 144 Tmth TWW for PndicaU Logic: FI

    CHAPTER SUMMARY EXERCISES Here are the new ideas from this chapter. Make sure you understand them, and record your summaries in your notebook. a) Rule V d) Rule -V b) Rule 3 e) Rule -3 c) New Name Requirement f) Reading an Interpretation Off an Open Branch


    3.7.4: Negated Quantified Sentences is shared under a not declared license and was authored, remixed, and/or curated by LibreTexts.

    • Was this article helpful?