1. Originally Posted by emakarov
First, in b), we don't have 1 in the universe; we have $\displaystyle \{1\}$, i.e., $\displaystyle \{\{\emptyset\}\}$. Recall that we are checking whether the formula $\displaystyle \forall x,y.\,x=y\leftrightarrow\forall z.\,z\in x\leftrightarrow z\in y$ is true in our universe. Therefore, $\displaystyle x$ and $\displaystyle y$ range over (i.e., take values from) our universe, which is $\displaystyle \{0,\{1\},2\}$, or $\displaystyle \{\emptyset,\{\{\emptyset\}\},\{\emptyset,\{\empty set\}\}\}$. So you should compare (as one of three pairs) $\displaystyle x=\emptyset$ and $\displaystyle y=\{\{\emptyset\}\}$.

I dont know how to write the symbols thats why I wrote 0 and 1. For x=$\displaystyle \emptyset$ and $\displaystyle y=\{\{\emptyset\}\}$ we have that for z=$\displaystyle \emptyset$ , z$\displaystyle \in\ emptyset$ and $\displaystyle z\notin \{\{\emptyset\}\}$. Correct ?

And for x=$\displaystyle \emptyset$ and $\displaystyle y =\{\emptyset\{\emptyset\}\}$ for z=$\displaystyle \emptyset$, $\displaystyle z\in\ emptyset \$ and z$\displaystyle \in \{\emptyset\{\emptyset\}\}$

Is that all I have to say for universe b ?

2. Originally Posted by emakarov

It seems to me that this moment is somewhat glossed over in the problem statement. That's why I said earlier: Here 0 is $\displaystyle \emptyset$ and 2 is $\displaystyle \{\emptyset,\{\emptyset\}\}$. You can write 2 differently, such as $\displaystyle \{\{\emptyset\},\emptyset,\emptyset\}$ (since the order and multiplicity of elements is ignored for sets), but there is no way to write 0 and 2 in the same way.

So they are the same. Since 0 (emptyset) is the "only" element that both have. Correct ? For 0 and 2 I talk for the first universe.

3. I dont know how to write the symbols thats why I wrote 0 and 1.
There is a big difference between 1 and {1}. 1 is an element of {1} but not the other way around. Also, the empty set is an element of 1 but not of {1}.

Let's agree to write 0 for the empty set $\displaystyle \emptyset$. All other sets can be written using curly braces and commas. You can also write "x in y" for $\displaystyle x\in y$.

For x= and we have that for z= , z and .
By definition, nothing is an element of the empty set. You keep repeating this mistake even after it was pointed out.

So they are the same. Since 0 (emptyset) is the "only" element that both have. Correct ? For 0 and 2 I talk for the first universe.
It seems that you don't understand the set notation. First, x is not the same as {x}. Also, I am not sure you understand the difference between an element and a subset.

4. Well I officially quit. I dont get it noway. If 0 and {0} -zero stands for emptyset- are not equal in the first universe, 0 and {0,{0}} are also not equal both in the first and in the second universe e.t.c. I cant figure out how to prove the type.

Thanks a lot for your time. I think you ' re right. I will fail logic. It is too difficult for me.

Thanks one more. You're very kind.

5. Sorry just one last question because I think I got it.

In the second universe {1} and 2 dont have the same elements but they are equal correct ??

empty set and one are IN {1}
empty set and one are IN 2

but their elements are different... ???

6. That HAS TO BE it. Thank you. I got it.

Thankssssssssssssssssssssss.

7. You are welcome.

Your misunderstanding so far may have come from something very basic, like the notation used to describe sets. A wrong understanding of this may render everything else incomprehensible. If this is the case, then your problems are understandable, and they do not mean that you have problems with logic in general.

In the second universe {1} and 2 dont have the same elements but they are equal correct ??

empty set and one are IN {1}
empty set and one are IN 2
No, {1}={{0}} and 2={0,{0}}={0,1} are non-equal sets because they are written in different ways.

Also, {1} and 2 have different elements. I wrote a table before that for each of the three elements of the universe in b) says whether this element belongs to {1} and to 2. In particular, the empty set is not an element of {{1}}: it is not listed explicitly. But the empty set is an element of 2.

I would recommend making sure you understand the notation used to describe sets as well as the concepts of an element and of a subset. You can read about them in a textbook or in Wikipedia. Nobody can move forward until these things are absolutely clear.

8. Originally Posted by emakarov
You are welcome.

No, {1}={{0}} and 2={0,{0}}={0,1} are non-equal sets because they are written in different ways.

Also, {1} and 2 have different elements.
Sorry {1} doesnt include (0,1)? It includes only number one? not zero?

From Wikipedia: (http://en.wikipedia.org/wiki/Natural_number)

2 = {1} = {{{ }}}, etc.

I'm sure now I got it. Maybe I dont express myself in a propre way but I think that's the solution.

9. prove by induction

Originally Posted by milagros
Hi. I have this problem to solve:

(∀x(P(x,x)∧∀x∀y∀z[(P(x,y)∧P(y,z))→P(x,z)]∧∀x∀y[P(x,y)∨P(y,x)])→∃y∀xP(y,x).

In natural language this means that if a relation P is reflexive and transitive and for two elements a,b we have P(a,b)∨P(b,a) then there is another element d in the universe that for every other element c of the universe we have P(d,c).
Sorry got stuck on how to prove by induction the above type for every finite set.

Ok for n=1 lets take universe A={a}. So:
case first : P={emptyset}
case second: P={(a,a)}
Lets say that its valid for n=k.
We ll prove for n=k+1 & we have universe:A={a0,a1,a2,...,a(k),a(k+1)}.
How do I go on?

Thanks in advance for any help.

10. Ok for n=1 lets take universe A={a}. So:
case first : P={emptyset}
case second: P={(a,a)}
Writing "case first" does not constitute a proof. When I grade(d) a homework, I am willing to allow sketchy proofs for someone who has consistently demonstrated that they understand the material and are more than able to fill in missing details. In this situation, however, the case P={emptyset} (which probably is supposed to mean P=emptyset) is impossible, which is not clear at all from your version.

We ll prove for n=k+1 & we have universe:A={a0,a1,a2,...,a(k),a(k+1)}.
How do I go on?
One can find the minimum of k+1 elements by finding the minimum m of the first k elements and then comparing m with the (k+1)st element.

11. Originally Posted by emakarov
Writing "case first" does not constitute a proof. When I grade(d) a homework, I am willing to allow sketchy proofs for someone who has consistently demonstrated that they understand the material and are more than able to fill in missing details. In this situation, however, the case P={emptyset} (which probably is supposed to mean P=emptyset) is impossible, which is not clear at all from your version.
Our professor wrote all this to help us make a begin. That's not my thought.
Originally Posted by emakarov
One can find the minimum of k+1 elements by finding the minimum m of the first k elements and then comparing m with the (k+1)st element.
something more specific on how to go on. I m so stupid with induction.

12. Proving ∃y∀xP(y,x) under assumption
(∀x(P(x,x)∧∀x∀y∀z[(P(x,y)∧P(y,z))→P(x,z)]∧∀x∀y[P(x,y)∨P(y,x)]) is equivalent to finding a minimum of the universe with respect to the reflexive, transitive and total relation P. I wrote that the problem of finding a minimum of k+1 elements can be reduced to finding a minimum of k elements and of 2 elements.

13. Originally Posted by emakarov
Proving ∃y∀xP(y,x) under assumption
(∀x(P(x,x)∧∀x∀y∀z[(P(x,y)∧P(y,z))→P(x,z)]∧∀x∀y[P(x,y)∨P(y,x)]) is equivalent to finding a minimum of the universe with respect to the reflexive, transitive and total relation P. I wrote that the problem of finding a minimum of k+1 elements can be reduced to finding a minimum of k elements and of 2 elements.
I think I got it. Thank you.

14. How can I prove the third relation??

P(x,y)\/P(y,x) ??

15. What do you mean?

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