24/03/09 19:56:37.16 P9QGsLiu.net
>>286
Next, we demonstrate that the existence of A and A having elements in space are equivalent.
Suppose there exists an entity A that exists but does not have elements in space.
Consider A and its power set {A}.
If we assume that {A} also does not have elements in space,
and define B as a set of entities that do not have elements in space. Considering {B}, if {B} is assumed to not have elements in space, then B⊇{B} and {B}⊇B would mean B and {B} coincide. However, this contradicts Lemma 1.1.
Therefore, {B} is an element of space. Thus, by Lemma 1.2, both A and {A} are shown to be elements of space, which contradicts the premise.
Hence, {A} is an element of space.
Furthermore, from Lemma 1.2, it can be deduced that A is also an element of space.
Therefore, if A exists, it is shown that A has elements in space.
Since it is self-evident that having elements in space implies existence, it is demonstrated that existence and having elements in space are equivalent. Furthermore, having elements in space is equivalent to having a contour, thus existing is equivalent to having a silhouette.