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Thread: Negate a statement

  1. #1
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    Negate a statement

    Exercise 1(e) in Section 2.2 of Daniel Velleman's How to prove it:

    Negate the statement and then reexpress the result as an equivalent positive statement:

    $\displaystyle A \cup B \subseteq C \backslash D$

    My attempt at the solution is by expressing it in the logical form as
    Step 1:$\displaystyle \forall x ((x \in A \vee x \in B) \Rightarrow (x \in C \wedge x \not \in D))$

    Step 2: negate the statement
    $\displaystyle \neg \forall x ((x \in A \vee x \in B) \Rightarrow (x \in C \wedge x \not \in D))$

    $\displaystyle \equiv \exists x \neg((x \in A \vee x \in B) \Rightarrow (x \in C \wedge x \not \in D))$

    $\displaystyle \equiv \exists x ((x \in A \vee x \in B) \wedge \neg (x \in C \wedge x \not \in D))$

    $\displaystyle \equiv \exists x ((x \in A \vee x \in B) \wedge \exists x (x \not \in C \vee x \in D))$

    $\displaystyle \equiv \exists x (x \in A \vee x \in B) \wedge \exists x (x \in C \Rightarrow x \in D)$

    Question: Is there a set notation that is equivalent to the last line?
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  2. #2
    MHF Contributor undefined's Avatar
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    I think your "exists x" inside parentheses in the last two steps are typos (or if not.. then they are mistakes).

    How about

    $\displaystyle \,|(A\cup B)\cap(C^c\cup D)|>0$

    ?
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  3. #3
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    Quote Originally Posted by undefined View Post
    I think your "exists x" inside parentheses in the last two steps are typos (or if not.. then they are mistakes).

    How about

    $\displaystyle \,|(A\cup B)\cap(C^c\cup D)|>0$

    ?
    No it's not typos. They violate the rule. Thanks for pointing it out.

    Will it be better that the expression be $\displaystyle (A \cup B) \cap (C^c \cup D) \not = \varnothing$?
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  4. #4
    MHF Contributor undefined's Avatar
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    Quote Originally Posted by novice View Post
    No it's not typos. They violate the rule. Thanks for pointing it out.

    Will it be better that the expression be $\displaystyle (A \cup B) \cap (C^c \cup D) \not = \varnothing$?
    I figured that wouldn't count as a positive statement.
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