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Math Help - 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:

    A \cup B \subseteq C \backslash D

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

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

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

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

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

    \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

    \,|(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

    \,|(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 (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 (A \cup B) \cap (C^c \cup D) \not = \varnothing?
    I figured that wouldn't count as a positive statement.
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