Negate a statement

**novice** · October 6th 2010, 07:41 PM

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?

**undefined** · October 6th 2010, 07:59 PM

I think your "exists x" inside parentheses in the last two steps are typos (or if not.. then they are mistakes).

How about

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

?

**novice** · October 7th 2010, 04:31 AM

Originally Posted by undefined

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 $(A \cup B) \cap (C^c \cup D) \not = \varnothing$ ?

**undefined** · October 7th 2010, 05:12 AM

Originally Posted by novice

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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