Set operations - is this a valid proof?

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Posted by Joel on October 01, 2002 at 19:11:12:

using the following symbols:
u = union
e = "is member of"
v = intersection
| = such that
c= = "is subset of"

is this a valid proof that
(A u B) c= (A u B u C) ("A union B is a subset of A union B union C")?

given an X | X e (A u B)
then (X e A) v (X e B)
then (X e A) v (X e B) v (X e C) [this is the step that worries me, since no information is given regarding membership or non-membership in C, but there has to be some way to introduce C into the picture]

then X e (A u B u C)
therefore any (X | X e (A u B)) e (A u B u B)
so by the definition of subset, (A u B) c= (A u B u C)

The whole question seems so obvious and trivial, and this argument seems so simple that it is hard to believe that this is a legitimate proof. Also, I'm bothered by the fact that "C" seems to materialize out of thin air, but there has to be some "legal" way to get C into the argument. Is this it?

Follow Ups:

Re: Set operations - CORRECTED - please look at this one instead Joel 19:14:45 10/01/02 (4)
- Re: Set operations - CORRECTED - please look at this one instead T.Gracken 07:57:31 10/02/02 (3)
  - Re: Set operations - CORRECTED - please look at this one instead Joel 10:06:30 10/02/02 (2)
    - not so... T.Gracken 11:55:25 10/02/02 (1)
      - You make it sound so easy. Thanks. Joel 22:50:31 10/02/02 (0)

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Comments:
: using the following symbols: : u = union : e = "is member of" : v = intersection : | = such that : c= = "is subset of" : is this a valid proof that : (A u B) c= (A u B u C) ("A union B is a subset of A union B union C")? : given an X | X e (A u B) : then (X e A) v (X e B) : then (X e A) v (X e B) v (X e C) [this is the step that worries me, since no information is given regarding membership or non-membership in C, but there has to be some way to introduce C into the picture] : then X e (A u B u C) : therefore any (X | X e (A u B)) e (A u B u B) : so by the definition of subset, (A u B) c= (A u B u C) : The whole question seems so obvious and trivial, and this argument seems so simple that it is hard to believe that this is a legitimate proof. Also, I'm bothered by the fact that "C" seems to materialize out of thin air, but there has to be some "legal" way to get C into the argument. Is this it?

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