5.10: Estructuras cubiertas para idiomas de primer orden
- Page ID
- 103689
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Recordemos que un término se cierra si no contiene variables.
Definición\(\PageIndex{1}\): Value of closed terms
Si\(t\) es un término cerrado del idioma\(\Lang L\) y\(\Struct M\) es una estructura para\(\Lang L\), el valor\(\Value{t}{M}\) se define de la siguiente manera:
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Si\(t\) es solo el símbolo constante\(c\), entonces\(\Value{c}{M} = \Assign{c}{M}\).
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Si\(t\) es de la forma\(\Atom{f}{t_1, \ldots, t_n}\), entonces\[\Value{t}{M} = \Assign{f}{M}(\Value{t_1}{M}, \ldots, \Value{t_n}{M}).\nonumber\]
Definición\(\PageIndex{2}\): Covered structure
Se cubre una estructura si cada elemento del dominio es el valor de algún término cerrado.
Ejemplo\(\PageIndex{1}\)
Dejar\(\Lang L\) ser el lenguaje con símbolos constantes\(\Obj{zero}\),,\(\Obj{one}\),...\(\Obj{two}\), el símbolo de predicado binario\(<\), y los símbolos de función binarios\(+\) y\(\times\). Entonces una estructura\(\Struct M\) para\(\Lang L\) es aquella con dominio\(\Domain M = \{0, 1, 2, \ldots \}\) y asignaciones\(\Assign{\Obj{zero}}{M} = 0\),\(\Assign{\Obj{one}}{M} = 1\),\(\Assign{\Obj{two}}{M} = 2\), y así sucesivamente. Para el símbolo de relación binaria\(<\), el conjunto\(\Assign{<}{M}\) es el conjunto de todos los pares\(\tuple{c_1, c_2} \in \Domain{M}^2\) tal que\(c_1\) es menor que\(c_2\): por ejemplo,\(\tuple{1, 3} \in \Assign{<}{M}\) pero\(\tuple{2, 2} \notin \Assign{<}{M}\). Para el símbolo de función binaria\(+\),\(\Assign{+}{M}\) defina de la manera habitual, por ejemplo,\(\Assign{+}{M}(2,3)\) mapea a\(5\), y de manera similar para el símbolo de función binaria\(\times\). Por lo tanto, el valor de\(\Obj{four}\) es justo\(4\), y el valor de\(\times(\Obj{two}, +(\Obj{three},\Obj{zero}))\) (o en notación infija,\(\Obj{two} \times (\Obj{three} + \Obj{zero})\)) es\[\begin{gathered} \begin{aligned} \Value{\times(\Obj{two}, +(\Obj{three},\Obj{zero})}{M} & =\Assign{\times}{M}(\Value{\Obj{two}}{M}, \Value{+(\Obj{three}, \Obj{zero})}{M})\\ & = \Assign{\times}{M}(\Value{\Obj{two}}{M}, \Assign{+}{M}(\Value{\Obj{three}}{M}, \Value{\Obj{zero}}{M})) \\ & = \Assign{\times}{M}(\Assign{\Obj{two}}{M}, \Assign{+}{M}(\Assign{\Obj{three}}{M}, \Assign{\Obj{zero}}{M})) \\ & = \Assign{\times}{M}(2, \Assign{+}{M}(3, 0)) \\ & = \Assign{\times}{M}(2, 3) \\ & = 6 \end{aligned}\end{gathered}\]
Problema\(\PageIndex{1}\)
¿Está cubierto\(\Struct N\), el modelo estándar de aritmética? Explicar.