1. ## 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?

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

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

?

3. 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).

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

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

Will it be better that the expression be $(A \cup B) \cap (C^c \cup D) \not = \varnothing$?

4. 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.