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279(1): ◆Ph05QxAcng [] 2024/03/09(土)19:31 ID:P9QGsLiu(11/21)
>>278
英訳
Theorem 2: Existence is a silhouette.
Proof
First, we will demonstrate two lemmas.
Lemma 2.1: Consider an element 'a' and its power set {a}. These two are different entities.
Proof
Assume they are the same.
If there exist elements a, b, then the set {a, b} would be the same entity. The latter is seen as a single entity consisting of a and b, which are clearly different. Thus, the initial premise is contradicted, proving the lemma.
Lemma 2.2: Elements of a set that are elements of space are themselves elements of space.
Proof
Let a set be E, and suppose an element e of E is not an element of space. If E∩e = e, then despite e being a part of E, there would exist a part that is not an element of space, which is a contradiction. Therefore, the lemma is proven.
Taking the contrapositive, we obtain the following corollary.
Corollary 2.2.1: A set composed of elements that are not elements of space is not an element of space.
280: ◆Ph05QxAcng [] 2024/03/09(土)19:31 ID:P9QGsLiu(12/21)
>>279
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 2.1.
Therefore, {B} is an element of space. Thus, by Lemma 2.2, 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 2.2, 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 defined contour.
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