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I want to ask that, are you empty sets equal ????
I reply, therefore I am not an empty set.I want to ask that, are you empty sets equal ????
You've defined two numbers, not two sets, but passing over that...How can two Empty Sets be equal ???
take a look at this example..
1st set: Number of trees in the forest
2nd Set: Number of Student of in the class
suppose that these are empty sets.. how are they equal ??? they both have different description..
I don't get it: you showed ##A\cup\ B=B\cup\ A## for A,B being (supposedly)The empty set is unique and the following proof ascertains that.
Suppose that there is another empty set denoted by ##\emptyset'## ,then we have:
1)##\forall A[A\cup\emptyset =A]##
2)##\forall A[A\cup\emptyset' =A]##
In (1) we put ##A =\emptyset'## and we get: ##\emptyset'\cup\emptyset =\emptyset'##
In (2) we put ## A =\emptyset## and we get :##\emptyset\cup\emptyset' =\emptyset##
But since ,##\emptyset'\cup\emptyset = \emptyset\cup\emptyset' ##
We can conclude :##\emptyset'=\emptyset ##
Hence all the empty sets are equal
WE have :I don't get it: you showed ##A\cup\ B=B\cup\ A## for A,B being (supposedly)
different copies of the empty set. But
this is true for any two sets, since union is commutative; I don't see how
this shows that A=B.
This is wrong as the followinf argument shows:. If two empty set were not equal, there would be one element in one of the sets which is not a member of the other set, and this is impossible, since an empty set has no member at all, whether member or not member of another empty set.
This argument is fallacious. You cannot infer that one and the same x is referred to by the two existential quantifers. If you could, it would be easy to prove that something is a Dodge and something is a Toyota implies that some one thing is both a Dodge and a Toyota. Maybe it would help to see this if you make the variable used in the second existential claim 'y'.##[\exists x(x\in \emptyset\wedge \neg x\in \emptyset')]\vee[\exists x(x\in \emptyset'\wedge \neg x\in \emptyset )]##.That implies :
##[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##.Which in turn implies:
##[ (x\in \emptyset\vee x\in \emptyset')]##
This argument is fallacious. You cannot infer that one and the same x is referred to by the two existential quantifers. If you could, it would be easy to prove that something is a Dodge and something is a Toyota implies that some one thing is both a Dodge and a Toyota. Maybe it would help to see this if you make the variable used in the second existential claim 'y'.
I am not saying that the use of the different quantifiers makes a difference, it does not. I am saying that Existential Elimination does not allow you to assume that the x referred to in the first existential claim is one and the same x as the x referred to in the second existential claim.Let : A= {1,2}, and B={1,2,3}.
Can you prove that :## A\neq B## , using the axiom of extensionality??.
Then you will find out that using the same or different quantifiers makes no difference
Your argument shows that I was right, but perhaps that is what you meant?This is wrong as the followinf argument shows:
To be clear, you should have written the two last lines as##[\exists x(x\in \emptyset\wedge \neg x\in \emptyset')]\vee[\exists x(x\in \emptyset'\wedge \neg x\in \emptyset )]##.That implies :
##[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##.Which in turn implies:
##[ (x\in \emptyset\vee x\in \emptyset')]##
Can you support that ,by writing a complete formal proof??Your argument shows that I was right, but perhaps that is what you meant?
To be clear, you should have written the two last lines as
##\exists x[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##
and
##\exists x[ (x\in \emptyset\vee x\in \emptyset')]##.
Then, the argument is correct, since
##\exists x(P(x)\vee Q(x))## and ##\exists x \,P(x)\vee\exists x\,Q(x)## are logically equivalent.
Oops, the parenteses here are partially wrong. It should be:##\exists x[ (x\in \emptyset\wedge \neg x\in \emptyset')]\vee[(x\in \emptyset'\wedge \neg x\in \emptyset )]##
Well, how did you get the corresponding line, in your previous post?Can you support that ,by writing a complete formal proof??
Because i know that it will be useless to ask you ,where did you get the:
"and ##\exists x[ (x\in \emptyset\vee x\in \emptyset')]##" ,part, e.t.c ,e.t.c
Write a formal proof supporting your argument in any system you like using any rules of inference you like ,i can follow.Well, how did you get the corresponding line, in your previous post?
If you want a complete formal proof, you must specify which formal system that should be used: which are the axioms and the rules of inference? Is a Hilbert style axiom system (and which variant in this case) or a natural deduction system (and which variant in this case) or some other kind of system?
And whatever system is used, complete formal proofs tend to be extremely lengthy. One almost always takes shortcuts. But you have a habit of questioning all possible shortcuts.
Of course there is. It's not because something is not written as a formal proof that you can't see whether it is right or wrong. In fact, when something is not written as a formal proof, it is much easier for me to grasp the proof.There is no other way of checking whether your argument is right or wrong.