# Truth table correct?

• Aug 8th 2007, 12:29 PM
shilz222
Truth table correct?
Is this truth table correct?

Thanks

Edit: should be $\displaystyle x \in A$, $\displaystyle x \in B$ etc...

For the last two I thought of them like this: If $\displaystyle x \in A, x \in B$ then $\displaystyle x \in A \cup B$ is true. However if $\displaystyle x \in C$ then $\displaystyle x \in (A \cup B) - C$ is false. So we require that $\displaystyle x \not \in C$ for the statement to be true. So the $\displaystyle '-'$ operation is sort of like the $\displaystyle \cap$ operation in terms of truths.

Crap...just realized I screwed up. For all $\displaystyle x \in C$ then the second to last statement is false. For all $\displaystyle x \not \in C$ then that statement is true.

The last one is basically a union of two sets. So to be false, $\displaystyle x \not \in A-B$ and $\displaystyle x \not \in C$
• Aug 8th 2007, 01:29 PM
Plato
Frankly, I am not at all sure what you are trying to do here.
Not sure how your text/instructor applies truth-values to sets.

However, here is the traditional understanding.
$\displaystyle \left( {A \cup B} \right) - C$ would be (A or B) and not C; $\displaystyle \left( {A \vee B} \right) \wedge (\neg C)$.

Whereas $\displaystyle \left( {A - B} \right) \cup C$ would be (A and not B) or C; $\displaystyle \left( {A \wedge \neg B} \right) \vee C$.

This is not what your table of values represents!
• Aug 8th 2007, 01:34 PM
shilz222
Thats why I added the $\displaystyle x \in A, x \in B \ldots$.

Otherwise they wouldn't be statements (i.e. you can't say the set $\displaystyle A$ is false).
• Aug 8th 2007, 01:50 PM
Plato
Quote:

Originally Posted by shilz222
Thats why I added the $\displaystyle x \in A, x \in B \ldots$. Otherwise they wouldn't be statements (i.e. you can't say the set $\displaystyle A$ is false).

But even with those additions, what in what sense does it make to assign a truth-value to $\displaystyle x \in A$?

Are you try to decide the truth-value of:
$\displaystyle \text{If}\;x \in A,\;x \in B\;\text{and}\;x \notin C\;\text{then}\;x \in \left( {A \cup B} \right) - C$?
• Aug 8th 2007, 01:55 PM
shilz222
No I am trying to do the following:

$\displaystyle x \in A, \ x \in B, \ x \in C, \ x \in (A \cup B) - C, \ x \in (A-B) \cup C$ are all statements.

Therefore we can assign truth values to them. So for the statement $\displaystyle x \in (A \cup B) - C$ to be true, we require that $\displaystyle x \not \in C$ or $\displaystyle x \in C$ to be false right? If $\displaystyle x \in C$ then the statement $\displaystyle x \in (A \cup B) - C$ is false (looking at a venn-diagram).

So the statement $\displaystyle x \in A$ is true means $\displaystyle x \in A$.

If the statement $\displaystyle x \in A$ is false then $\displaystyle x \not \in A$.
• Aug 8th 2007, 01:57 PM
Plato
Well I would expect that it means this:
• Aug 8th 2007, 02:01 PM
shilz222
Yes that what I meant in my first post (with the $\displaystyle x \in A$ $\displaystyle x \in B$ etc..).

Thanks