8.9: Derivabilidad y consistencia
- Page ID
- 103665
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Ahora estableceremos una serie de propiedades de la relación de derivabilidad. Son independientemente interesantes, pero cada uno jugará un papel en la prueba del teorema de integridad.
Si\(\Gamma \Proves A\) y\(\Gamma \cup \{A\}\) es inconsistente, entonces\(\Gamma\) es inconsistente.
Prueba. Hay finitos\(\Gamma_0\) y\(\Gamma_1 \subseteq \Gamma\) tales que\(\Log{LK}\) deriva\(\Gamma_0 \Sequent A\) y\(A, \Gamma_1 \Sequent \quad\). Que la\(\Log{LK}\) -derivación de\(\Gamma_0 \Sequent A\) ser\(\pi_0\) y la\(\Log{LK}\) -derivación de\(\Gamma_1, A \Sequent \quad\) ser\(\pi_1\). Entonces podemos derivar
Desde\(\Gamma_0 \subseteq \Gamma\) y\(\Gamma_1 \subseteq \Gamma\),\(\Gamma_0 \cup \Gamma_1 \subseteq \Gamma\), por lo tanto\(\Gamma\) es inconsistente. ◻
\(\Gamma \Proves A\)iff\(\Gamma \cup \{\lnot A\}\) es inconsistente.
Prueba. Primero supongamos\(\Gamma \Proves A\), es decir, hay una derivación\(\pi_0\) de\(\Gamma \Sequent A\). Al agregar una\(\LeftR{\lnot}\) regla, obtenemos una derivación de\(\lnot A, \Gamma \Sequent \quad\), es decir,\(\Gamma \cup \{\lnot A\}\) es inconsistente.
Si\(\Gamma \cup \{\lnot A\}\) es inconsistente, hay una derivación\(\pi_1\) de\(\lnot A, \Gamma \Sequent \quad\). La siguiente es una derivación de\(\Gamma \Sequent A\):
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Problema\(\PageIndex{1}\)
Demostrar que\(\Gamma \Proves \lnot A\) iff\(\Gamma \cup \{A\}\) es inconsistente.
Si\(\Gamma \Proves A\) y\(\lnot A \in \Gamma\), entonces\(\Gamma\) es inconsistente.
Prueba. Supongamos\(\Gamma \Proves A\) y\(\lnot A \in \Gamma\). Después hay una derivación\(\pi\) de un secuente\(\Gamma_0 \Sequent A\). El secuente también\(\lnot A, \Gamma_0 \Sequent \quad\) es derivable:
Desde\(\lnot A \in \Gamma\) y\(\Gamma_0 \subseteq \Gamma\), esto demuestra que\(\Gamma\) es inconsistente. ◻
Si\(\Gamma \cup \{A\}\) y ambos\(\Gamma \cup \{\lnot A\}\) son inconsistentes, entonces\(\Gamma\) es inconsistente.
Prueba. Hay conjuntos finitos\(\Gamma_0 \subseteq \Gamma\) y\(\Gamma_1 \subseteq \Gamma\) y\(\Log{LK}\) -derivaciones\(\pi_0\) y\(\pi_1\) de\(A, \Gamma_0 \Sequent \quad\) y\(\lnot A, \Gamma_1 \Sequent \quad\), respectivamente. Entonces podemos derivar
Desde\(\Gamma_0 \subseteq \Gamma\) y\(\Gamma_1 \subseteq \Gamma\),\(\Gamma_0 \cup \Gamma_1 \subseteq \Gamma\). De ahí\(\Gamma\) que sea inconsistente. ◻