Re: Set Theory (help again)

For any propositions A and B, I'll write ~A to mean the negation of A and A => B to mean "A implies B." Also, If P is a property (or predicate, such as "even" or "mortal"), I'll write P(x) to mean that x satisfies P. Therefore, ~P(x) means that x does not satisfy P.

If A and B are propositions, then ~A => ~B does not in general imply A => B. However, ~A => ~B is equivalent to B => A. The formula ~A => ~B is called the contrapositive of B => A.

Question A says "For every x, ~A(x) => ~B(x)." Using the information above, this is the same as "For every x, B(x) => A(x)," but not the same as "For every x, A(x) => B(x)."

B is in fact false. If X, Y and Z are propositions, then (X or Y) => Z is equivalent ((X => Z) and (Y => Z)), but x ∈ A ∩ B does not imply that x ∈ A ∩ C.

E is true: consider two cases, when x ∈ A and x ∈ B.

G is false: consider x ∈ B, but x ∉ A and x ∉ C.

I'll let someone else to do the rest, or I'll return to this later.

Re: Set Theory (help again)

Thanks again!

Wouldn't G be True?

If x is in A∪B or x is in C then x is in A∪C and x is in B∪C.

Since there is an 'or', only one of the condition needs to be met correct? That being in the case, we only use "x is in C" and not "x is in A∪B":

When x is in C, then x is in A∪C as well as in B∪C right? I'm not sure if I'm misunderstanding anything though :(.

Also are A and F False?

Re: Set Theory (help again)

Quote:

Originally Posted by

**PennCow** Wouldn't G be True?

If x is in A∪B or x is in C then x is in A∪C and x is in B∪C.

Since there is an 'or', only one of the condition needs to be met correct? That being in the case, we only use "x is in C" and not "x is in A∪B"

First, I stand by my suggestion to consider the situation where x ∈ B, but x ∉ A and x ∉ C. Second, as I said with respect to question B, (X or Y) => Z is equivalent ((X => Z) and (Y => Z)). When you are proving (X or Y) => Z, you assume X or Y and need to show Z. You don't know which of X and Y is true, only that one of them is. Therefore, you need to show Z assuming just X, *and* also show Z assuming just Y. In question G, you don't have the right to assume that the premise (x ∈ A ∪ B or x ∈ C) means x ∈ C. Instead, you need to show that x ∈ A ∪ C assuming just x ∈ A ∪ B and also show x ∈ A ∪ C assuming just x ∈ C. The first of these two implications is false in general.

Quote:

Originally Posted by

**PennCow** Also are A and F False?

As I said, Quote:

Originally Posted by

**emakarov** Question A says "For every x, ~A(x) => ~B(x)." Using the information above, this is the same as "For every x, B(x) => A(x)," but not the same as "For every x, A(x) => B(x)."

Re: Set Theory (help again)

Ahh I see, thanks. Been up all night so I think I am misreading your stuff as we go... gotcha now >_<! Sorry!