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| Таким образом, <tex>\Sigma_{i} \subset \Sigma_{i+1}, \Sigma_{i} \subset \Pi_{i+1} \Rightarrow \Sigma_{i} \subset \Sigma_{i+1} \cap \Pi_{i+1}</tex>. | | Таким образом, <tex>\Sigma_{i} \subset \Sigma_{i+1}, \Sigma_{i} \subset \Pi_{i+1} \Rightarrow \Sigma_{i} \subset \Sigma_{i+1} \cap \Pi_{i+1}</tex>. |
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− | {{Теорема
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− | |statement = <tex>\Pi_{i} \subset \Sigma_{i+1} \cap \Pi_{i+1}</tex>.
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− | |proof = Пусть <tex>L \in \Pi_{i} \Rightarrow \exists R : x \in L \Leftrightarrow \forall y_{1} \cdots Q y_{i} : R(x,y_{1},\cdots,y_{i}), \forall j |y_{j}| \le poly(|x|)</tex>.<br/>
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− | Проверим, что <tex>L \in \Pi_{i+1} \Leftrightarrow \exists R' : x \in L \Leftrightarrow \forall y_{1} \cdots Q y_{i} \bar{Q} y_{i+1} : R'(x,y_{1},\cdots,y_{i},y_{i+1})</tex>.
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− | <br/>
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− | <tex>R'(x,y_{1},\cdots,y_{i+1})</tex> {
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− | return <tex>R(x,y_{1},\cdots,y_{i})</tex>;
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− | }
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− | Проверим, что <tex>L \in \Sigma_{i+1} \Leftrightarrow \exists R'' : x \in L \Leftrightarrow \exists y_{0} \forall y_{1} \cdots Q y_{i} : R''(x,y_{0},y_{1},\cdots,y_{i})</tex>.
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− | <br/>
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− | <tex>R''(x,y_{0},y_{1},\cdots,y_{i})</tex> {
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− | return <tex>R(x,y_{1},\cdots,y_{i})</tex>;
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− | }
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− | Таким образом, <tex>\Pi_{i} \subset \Sigma_{i+1}, \Pi_{i} \subset \Pi_{i+1} \Rightarrow \Pi_{i} \subset \Sigma_{i+1} \cap \Pi_{i+1}</tex>.
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| }} | | }} |
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Строка 39: |
| {{Определение | | {{Определение |
| |definition = | | |definition = |
− | <tex>\mathrm{PH_{1}} = {\bigcup \atop {i \in \mathbb{N}}} \Sigma_{i}</tex>.<br/> | + | <tex>\mathrm{PH} = {\bigcup \atop {i \in \mathbb{N}}} \Sigma_{i}</tex>.<br/> |
− | <tex>\mathrm{PH_{2}} = {\bigcup \atop {i \in \mathbb{N}}} \Pi_{i}</tex>.<br/>
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− | <tex>\mathrm{PH_{3}} = {\bigcup \atop {i \in \mathbb{N}}} (\Sigma_{i} \cup \Pi_{i})</tex>.
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− | }}
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− | | |
− | {{Теорема
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− | |statement = Все три определения класса <tex>PH</tex> эквивалентны, т.е. <tex>\mathrm{PH_{1}} = \mathrm{PH_{2}} = \mathrm{PH_{3}}</tex>.
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− | |proof = <tex>\Sigma_{i} \subset \Pi_{i+1} \Rightarrow \mathrm{PH_{1}} \subset \mathrm{PH_{2}}</tex>.<br/>
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− | <tex>\Pi_{i} \subset (\Sigma_{i+1} \cap \Pi_{i+1}) \subset (\Sigma_{i+1} \cup \Pi_{i+1}) \Rightarrow \mathrm{PH_{2}} \subset \mathrm{PH_{3}}</tex>.<br/>
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− | <tex>\Pi_{i} \subset \Sigma_{i+1}, \Sigma_{i} \subset \Sigma_{i+1} \Rightarrow \mathrm{PH_{3}} \subset \mathrm{PH_{1}}</tex>.<br/>
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− | Таким образом, <tex>\mathrm{PH_{1}} \subset \mathrm{PH_{2}} \subset \mathrm{PH_{3}} \subset \mathrm{PH_{1}}</tex>.
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| }} | | }} |
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