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293(1): ◆Ph05QxAcng 2024/03/10(日) 01:05:24.51 ID:86TMEkfZ(1/15)調 AAS
>>292
英訳です
Proposition: For every computational problem, there exists an algorithm that matches the class of computational complexity of that problem.
Proof:
First, we demonstrate the following theorem.
Theorem: The proof of every proposition that can be judged as true or false will inevitably materialize.
Proof:
First, we present three lemmas.
Lemma 1: The condition that defines an outline is when the inclusion relation ⊇ is commutative.
Proof:
If there is an A with a known shape and an A with an unknown shape, then the shape of B can be determined by showing A⊇B and B⊇A.
Corollary 1.1: The concept of existence and the inclusion relation are equivalent.
Proof:
The concept of existence and space are equivalent. Clearly, the inclusion relation is derived from space, and from the inclusion relation, existence=space is derived.
294(1): ◆Ph05QxAcng 2024/03/10(日) 01:05:56.34 ID:86TMEkfZ(2/15)調 AAS
>>293
Lemma 2: The inclusion relation ⊇ and the correspondence relation → are equivalent.
Proof:
If the inclusion relation A⊇B holds, then the correspondence relation A→B holds. Conversely, assume the correspondence relation A→B holds. In this case, the premise that if something exists, then all elements are in space, and from Corollary 1.1, that the existence of elements in space implies an inclusion relation, demonstrates the proposition.
Lemma 3: The inclusion relation ⊇ and the causal relation → are equivalent.
Proof:
Clearly, if a causal relation exists, then, as per Corollary 1.1, since there are elements in space, an inclusion relation is derived. Next, we demonstrate that a causal relation can be derived from an inclusion relation. When A⊇B holds, a causal relation that if B then A belongs, holds. Thus, the proposition is demonstrated.
Next, we provide the proof of the theorem.
Existence and outline are equivalent, outline and wave (inclusion relation ⊇) are equivalent, and the inclusion relation and causal relation (logic →) are equivalent. Therefore, taking the physical space of this visible reality = existence as the foremost premise, everything that follows is represented. In other words, it may be said that the essence of this world is the visible space. Therefore, if the proof of every true proposition exists, it contradicts not existing in visible space. Thus, the theorem is demonstrated.
Corollary: For every true proof problem, as time is infinitely increased (with ∞^n, where n can be made as large as possible), the ratio of solved problems converges to 1.
295(1): ◆Ph05QxAcng 2024/03/10(日) 01:06:26.84 ID:86TMEkfZ(3/15)調 AAS
>>294
Now, if we assume that the proposition does not hold, then for every problem, there exists a problem a whose algorithm does not match the class of the problem, and is of a smaller class.
At this time, it is possible to create a program A that generates problem a infinitely.
Let there be a program X that creates problems infinitely, and let r(X) be the ratio of the total problems solved by it.
In the case of A, even if we take n as large as possible with time being ∞^n, r(A) converges to 0.
Let P be the set of all true proof problems, and let p be the set of all computational problems, then P ⊃ p.
To solve a computational problem means to demonstrate the following two points:
(1) Decide whether there is a better algorithm than brute force search, and if it exists, demonstrate that it is the best; if not, demonstrate its non-existence.
(2) If it exists, verify whether a solution exists using that algorithm and specifically output the value. If it does not exist, verify the existence of a solution through brute force search.
296: ◆Ph05QxAcng 2024/03/10(日) 01:06:54.56 ID:86TMEkfZ(4/15)調 AAS
>>295
Let E be the set of numerous problems created by A. Since the corollary indicates that r converges to 1, both (1) and (2) are demonstrated for E and converge to 1 ⇔ for each problem, as all algorithms are the same, the best algorithm for a computational problem must match its class. However, as demonstrated earlier, r(A) converges to 0, which is a contradiction.
Thus, the proposition is demonstrated.
Corollary: P=NP holds.
297(1): ◆Ph05QxAcng 2024/03/10(日) 01:23:34.39 ID:86TMEkfZ(5/15)調 AAS
viXraに以下の論文を投稿しました。存在が輪郭である事の証明もつけてます。
A proof of P=NP
Sakura Tachibana
Abstract
I prove that existence is a silhouette. Next I prove P=NP.
Theorem1
Existence is a silhouette.
Proof
First, we will demonstrate two lemmas.
Lemma 1.1: 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 1.2: 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 1.2.1: A set composed of elements that are not elements of space is not an element of space.
298: ◆Ph05QxAcng 2024/03/10(日) 01:24:02.81 ID:86TMEkfZ(6/15)調 AAS
>>297
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.
299(1): ◆Ph05QxAcng 2024/03/10(日) 01:24:37.02 ID:86TMEkfZ(7/15)調 AAS
Next, I prove P=NP.
Proposition: For every computational problem, there exists an algorithm that matches the class of computational complexity of that problem.
Proof:
First, we demonstrate the following theorem.
Theorem2: The proof of every proposition that can be judged as true or false will inevitably materialize.
Proof:
First, we present three lemmas.
Lemma 2.1: The condition that defines an outline is when the inclusion relation ⊇ is commutative.
Proof:
If there is an A with a known shape and an A with an unknown shape, then the shape of B can be determined by showing A⊇B and B⊇A.
Corollary 2.1.1: The concept of existence and the inclusion relation are equivalent.
Proof:
The concept of existence and space are equivalent. Clearly, the inclusion relation is derived from space, and from the inclusion relation, existence=space is derived.
Lemma 2.2: The inclusion relation ⊇ and the correspondence relation → are equivalent.
Proof:
If the inclusion relation A⊇B holds, then the correspondence relation A→B holds. Conversely, assume the correspondence relation A→B holds. In this case, the premise that if something exists, then all elements are in space, and from Corollary 1.1, that the existence of elements in space implies an inclusion relation, demonstrates the proposition.
Lemma 2.3: The inclusion relation ⊇ and the causal relation → are equivalent.
Proof:
Clearly, if a causal relation exists, then, as per Corollary 1.1, since there are elements in space, an inclusion relation is derived. Next, we demonstrate that a causal relation can be derived from an inclusion relation. When A⊇B holds, a causal relation that if B then A belongs, holds. Thus, the proposition is demonstrated.
300(1): ◆Ph05QxAcng 2024/03/10(日) 01:25:08.80 ID:86TMEkfZ(8/15)調 AAS
>>299
Next, we provide the proof of the theorem2.
Existence and outline are equivalent, outline and wave (inclusion relation ⊇) are equivalent, and the inclusion relation and causal relation (logic →) are equivalent. Therefore, taking the physical space of this visible reality = existence as the foremost premise, everything that follows is represented. In other words, it may be said that the essence of this world is the visible space. Therefore, if the proof of every true proposition exists, it contradicts not existing in visible space. Thus, the theorem is demonstrated.
Corollary: For every true proof problem, as time is infinitely increased (with ∞^n, where n can be made as large as possible), the ratio of solved problems converges to 1.
Now, if we assume that the proposition does not hold, then for every problem, there exists a problem a whose algorithm does not match the class of the problem, and is of a smaller class.
At this time, it is possible to create a program A that generates problem a infinitely.
Let there be a program X that creates problems infinitely, and let r(X) be the ratio of the total problems solved by it.
In the case of A, even if we take n as large as possible with time being ∞^n, r(A) converges to 0.
301: ◆Ph05QxAcng 2024/03/10(日) 01:25:38.05 ID:86TMEkfZ(9/15)調 AAS
>>300
Let P be the set of all true proof problems, and let p be the set of all computational problems, then P ⊃ p.
To solve a computational problem means to demonstrate the following two points:
(1) Decide whether there is a better algorithm than brute force search, and if it exists, demonstrate that it is the best; if not, demonstrate its non-existence.
(2) If it exists, verify whether a solution exists using that algorithm and specifically output the value. If it does not exist, verify the existence of a solution through brute force search.
Let E be the set of numerous problems created by A. Since the corollary indicates that r converges to 1, both (1) and (2) are demonstrated for E and converge to 1 ⇔ for each problem, as all algorithms are the same, the best algorithm for a computational problem must match its class. However, as demonstrated earlier, r(A) converges to 0, which is a contradiction.
Thus, the proposition is demonstrated.
Corollary: P=NP holds.
302: ◆Ph05QxAcng 2024/03/10(日) 01:39:39.19 ID:86TMEkfZ(10/15)調 AAS
某有名数学者にメールを送りました
303(1): ◆Ph05QxAcng 2024/03/10(日) 02:46:01.53 ID:86TMEkfZ(11/15)調 AAS
この内容の論文をviXraに投稿しました。次に有名数学者に送ってタイムスタンプを付けます。
Fermat’s conjecture is true
Sakura Tachibana
Abstract
I prove that lines do not exist and addition does not hold. Thus Fermat’s conjecture is true.
Theorem 1: Lines do not exist.
Proof:
Consider the operation of creating numbers from {0,1} and the operations {+,-,*,/}. In this case, the sign of 1/0 is ±, and + and - intersect. The condition that curvature 0 exists ⟺ lines exist is contradictory because + and - intersect, so the condition cannot be added. Therefore, lines do not exist.
Theorem 2: All spatial structures that include lines do not exist.
Proof:
Self-evident
Theorem 3: Addition does not hold.
Proof:
Consider a vector space V constructed from {a}. In this case, -a∈V, but a, 0, and -a line up on a line, which contradicts Theorem 1.
Therefore, the theorem is proven.
Corollary 3.1: 0 does not exist.
Corollary 3.2: All propositions and proofs that assume addition and lines are false.
Corollary 3.3: Fermat’s conjecture is true.
Proof:
Because addition does not hold, so we can not define addition.
304: ◆Ph05QxAcng 2024/03/10(日) 02:47:59.64 ID:86TMEkfZ(12/15)調 AAS
>>303
送りました
305: ◆Ph05QxAcng 2024/03/10(日) 03:02:34.56 ID:86TMEkfZ(13/15)調 AAS
リーマン予想もviXraに提出しました
Riemann hypothesis is false
Sakura Tachibana
Abstract
I prove that lines do not exist and addition does not hold. Thus Riemann hypothesis is false.
Theorem 1: Lines do not exist.
Proof:
Consider the operation of creating numbers from {0,1} and the operations {+,-,*,/}. In this case, the sign of 1/0 is ±, and + and - intersect. The condition that curvature 0 exists ⟺ lines exist is contradictory because + and - intersect, so the condition cannot be added. Therefore, lines do not exist.
Theorem 2: All spatial structures that include lines do not exist.
Proof:
Self-evident
Theorem 3: Addition does not hold.
Proof:
Consider a vector space V constructed from {a}. In this case, -a∈V, but a, 0, and -a line up on a line, which contradicts Theorem 1.
Therefore, the theorem is proven.
Corollary 3.1: 0 does not exist.
Corollary 3.2: All propositions and proofs that assume addition and lines are false.
Corollary 3.3: Riemann hypothesis is false.
Proof:
Because lines do not exist, so Riemann hypothesis is false.
306: ◆Ph05QxAcng 2024/03/10(日) 03:21:11.48 ID:86TMEkfZ(14/15)調 AAS
BSD予想もサブミットして数学者に送りました
307: ◆Ph05QxAcng 2024/03/10(日) 04:22:55.77 ID:86TMEkfZ(15/15)調 AAS
ミレニアム懸賞問題全てとフェルマー予想、ビール予想、ゴールドバッハ予想のサブミットと有名数学者にメールを全て送りました、全員別の人です。
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