Let be non empty families of sets.
I am trying to prove that
I needed to prove something else. I had to come up with some counterexample which gave me the idea
for the above theorem. I am trying to see if I can prove it. Following is my try.
Since this is an implication , Givens are
and the Goal is
So the givens are , in logical language ,
Now, I put the goal in logical language. I am not very sure of this, please check it
since the goal is of the form , I let z be arbitrary and suppose that
So the new list of the givens is
and the goal now becomes
again , since the goal is of the form , I let t be arbitrary and suppose that
Finally the givens are
and the goal is simply
So after breaking this logically , we can now see from the givens that we let y be z and
since , we can conclude that
since z and t are arbitrary , the result holds for all values of z and t. But I am not
yet sure if this proves that
I need this for some related problem in Velleman's book "How to prove it". Author wants the
readers to explore all the logical structure of the proof. Hence all the details.
Thanks Plato. I had taken some other example. So obviously , the proposition tried
by me is wrong. But consider following families
here we have
so we have
in your case you got
So is there something wrong in my proof ?
Edit: so if
does it mean that in such case ?
Therefore, there is no proof.
What don't you understand about that?
All you did is to give an example in which it is true.
But we cannot prove by example. We disprove by counterexample.
The important thing is that your second goal means , i.e., ; it does not guarantee the converse inclusion.
You probably instantiate y with t, not z.Finally the givens are
and the goal is simply
So after breaking this logically , we can now see from the givens that we let y be z and since , we can conclude that
Otherwise, your proof is correct.
I am no way trying to pick an argument with this.
But I I do not understand why one would proceed to considering a proof if the OP is completely false. I have the same argument with this tread.
If the original posted question is false, why proceed to any consideration of a proof which cannot exist.
Now it may be an interesting logical exercise in the improper usage in incorrect instantiation. But from a purely mathematical point of view that is nonsense.
. and the condition for this is that there should be some . And from my list of givens, I see that there is . So I am instantiating y with t. Since we satisfy the conditions in the antecedent we proceed to the consequent
part. That part then guarantees that , y so chosen , is not in B. Since another given says that
the consequent part applies to this specific z , which is in G. Since y is instantiated with t , we conclude that .
So why do you think I am instantiating y with z ?
Anyway, as emakarov mentioned, even instantiating to t, the actual fatal error in your argument is the that you still didn't prove that 0 in F/\G. Your error was in misstating the goal. You stated only one direction of inclusion, and forgot the other direction.
It seems that finding an error in a proof serves the same goal as studying a correct proof. A person may accept on faith that a proof in the textbook is correct, but studying proofs will teach him/her to make their own proofs and will give a satisfaction that comes from a thoroughly understood subject matter. Similarly, finding an error rather than just accepting that it exists hopefully will teach how to avoid errors in the future and give a satisfaction that everything is clearly understood.