Recent content by xxxx0xxxx

Ok, if you want to be picky :), the definition leaves open the possibility that some elements of the relation ##R## are not ordered pairs taken from ##A## and ##B##.

Folks, here is the standard set theoretic definition of a relation: $$R~\text{ is a relation}~\Leftrightarrow~\forall r(r \in R~\rightarrow~\exists x \exists y(r~=~<x,y>)~)$$ This definition leave open the possibility that some members of relation ##R## are not ordered pairs. However, the...

If it's raining then it's cloudy, so either it's not raining or it's cloudy. $$p~{\rightarrow}~q~{\equiv}~{\neg} p~{\vee}~q$$ Suppose it's sunny, then it's not raining, and so either it's not sunny or it's not raining $${\neg} q~{\rightarrow}~{\neg} p~{\equiv}~q~{\vee}~{\neg}p$$

Because you statement says: for all y, if y is Jack, then Jack is Jill, whereas the correct statement says for all y and for all z, if y is Jack and z is Jack then Jack is Jack.

No it wouldn't matter, but the first choice is conventional. Pretty much from Suppes (1960)

Um yeah ...

You're right, but notationally, the elements need not be distinct. The definition of set makes them distinct, regardless of notation. The question posed is "why do the elements of a set have to be distinct?" The answer is "You are asking the wrong question." He will come across...

It is a consequence of the Pairing and Extensionality Axioms, for instance: \{ x, y, z \} = \{ x,y,y,z \}= \{ x,y,y,y,z \}= \{ x,\mbox{ ... any number of y's ... },z \}

Well, we ain't supposed to be doing homework problems here, but the proofs are extremely simple. for proof 1: Your notation is inconsistent, so can't help until you clean that up. for proof 2: \forall A (H(A,c) \Rightarrow H(P(A),c)) Assuming A, H(A,c), and P(A) exist immediately...

Neither one is better than the other, nor is one preferable to the other.

I go with: R \mbox{ is a relation} \Leftrightarrow \forall r(r \in R \Rightarrow \exists x \exists y (r=<x,y>)) This definition states that the elements of R are ordered pairs if and only if R is a relation.

Sorry Q, but you've only stated half of the equivalence, the other half does not follow..., unless you us a=a, that is, you have assumed the consequent.

I think it would be better to avoid all of the complexity, and use something simple like the following set theorem, which makes rich use of proof by contradiction: \forall x (x \not \in \emptyset)

Both are true, in the one case, x \mbox{ is a set} \Rightarrow ( \emptyset \subseteq x) and the other, x \in \{ \emptyset \} \Leftrightarrow x = \emptyset .

I usually observe that: (P \Rightarrow (P \wedge \neg P)) \Rightarrow \neg P is a tautology irrespective of any of the various set theories, and go from there.