ミレニアム懸賞問題 (635ﾚｽ)

ミレニアム懸賞問題 http://rio2016.5ch.net/test/read.cgi/math/1668766352/

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293: ◆Ph05QxAcng [] 2024/03/10(日) 01:05:24.51 ID:86TMEkfZ >>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. http://rio2016.5ch.net/test/read.cgi/math/1668766352/293

294: ◆Ph05QxAcng [] 2024/03/10(日) 01:05:56.34 ID:86TMEkfZ >>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. http://rio2016.5ch.net/test/read.cgi/math/1668766352/294

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