9.12: Derivaciones con predicado de identidad
- Page ID
- 103641
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\(\newcommand{\avec}{\mathbf a}\) \(\newcommand{\bvec}{\mathbf b}\) \(\newcommand{\cvec}{\mathbf c}\) \(\newcommand{\dvec}{\mathbf d}\) \(\newcommand{\dtil}{\widetilde{\mathbf d}}\) \(\newcommand{\evec}{\mathbf e}\) \(\newcommand{\fvec}{\mathbf f}\) \(\newcommand{\nvec}{\mathbf n}\) \(\newcommand{\pvec}{\mathbf p}\) \(\newcommand{\qvec}{\mathbf q}\) \(\newcommand{\svec}{\mathbf s}\) \(\newcommand{\tvec}{\mathbf t}\) \(\newcommand{\uvec}{\mathbf u}\) \(\newcommand{\vvec}{\mathbf v}\) \(\newcommand{\wvec}{\mathbf w}\) \(\newcommand{\xvec}{\mathbf x}\) \(\newcommand{\yvec}{\mathbf y}\) \(\newcommand{\zvec}{\mathbf z}\) \(\newcommand{\rvec}{\mathbf r}\) \(\newcommand{\mvec}{\mathbf m}\) \(\newcommand{\zerovec}{\mathbf 0}\) \(\newcommand{\onevec}{\mathbf 1}\) \(\newcommand{\real}{\mathbb R}\) \(\newcommand{\twovec}[2]{\left[\begin{array}{r}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\ctwovec}[2]{\left[\begin{array}{c}#1 \\ #2 \end{array}\right]}\) \(\newcommand{\threevec}[3]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\cthreevec}[3]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \end{array}\right]}\) \(\newcommand{\fourvec}[4]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\cfourvec}[4]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \end{array}\right]}\) \(\newcommand{\fivevec}[5]{\left[\begin{array}{r}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\cfivevec}[5]{\left[\begin{array}{c}#1 \\ #2 \\ #3 \\ #4 \\ #5 \\ \end{array}\right]}\) \(\newcommand{\mattwo}[4]{\left[\begin{array}{rr}#1 \amp #2 \\ #3 \amp #4 \\ \end{array}\right]}\) \(\newcommand{\laspan}[1]{\text{Span}\{#1\}}\) \(\newcommand{\bcal}{\cal B}\) \(\newcommand{\ccal}{\cal C}\) \(\newcommand{\scal}{\cal S}\) \(\newcommand{\wcal}{\cal W}\) \(\newcommand{\ecal}{\cal E}\) \(\newcommand{\coords}[2]{\left\{#1\right\}_{#2}}\) \(\newcommand{\gray}[1]{\color{gray}{#1}}\) \(\newcommand{\lgray}[1]{\color{lightgray}{#1}}\) \(\newcommand{\rank}{\operatorname{rank}}\) \(\newcommand{\row}{\text{Row}}\) \(\newcommand{\col}{\text{Col}}\) \(\renewcommand{\row}{\text{Row}}\) \(\newcommand{\nul}{\text{Nul}}\) \(\newcommand{\var}{\text{Var}}\) \(\newcommand{\corr}{\text{corr}}\) \(\newcommand{\len}[1]{\left|#1\right|}\) \(\newcommand{\bbar}{\overline{\bvec}}\) \(\newcommand{\bhat}{\widehat{\bvec}}\) \(\newcommand{\bperp}{\bvec^\perp}\) \(\newcommand{\xhat}{\widehat{\xvec}}\) \(\newcommand{\vhat}{\widehat{\vvec}}\) \(\newcommand{\uhat}{\widehat{\uvec}}\) \(\newcommand{\what}{\widehat{\wvec}}\) \(\newcommand{\Sighat}{\widehat{\Sigma}}\) \(\newcommand{\lt}{<}\) \(\newcommand{\gt}{>}\) \(\newcommand{\amp}{&}\) \(\definecolor{fillinmathshade}{gray}{0.9}\)Las derivaciones con predicado de identidad requieren reglas de inferencia adicionales.
En las reglas anteriores,\(t\),\(t_1\), y\(t_2\) son términos cerrados. La\(\Intro{\eq[][]}\) regla nos permite derivar cualquier declaración de identidad de la forma de\(\eq[t][t]\) plano, a partir de ninguna suposición.
Ejemplo\(\PageIndex{1}\)
Si\(s\) y\(t\) son términos cerrados, entonces\(A(s), \eq[s][t] \Proves A(t)\):
Esto puede ser familiar como el “principio de sustituibilidad de las mismas”, o Ley de Leibniz.
Problema\(\PageIndex{1}\)
Demostrar que\(=\) es tanto simétrico como transitivo, es decir, dar derivaciones de\(\lforall{x}{\lforall{y}{(\eq[x][y] \lif \eq[y][x])}}\) y\(\lforall{x}{\lforall{y}{\lforall{z}{}((\eq[x][y] \land \eq[y][z]) \lif \eq[x][z])}}\)
Ejemplo\(\PageIndex{2}\)
Derivamos la sentencia
\[\lforall{x}{\lforall{y}{((A(x) \land A(y)) \lif \eq[x][y])}} \nonumber\]
de la sentencia
\[\lexists{x}{\lforall{y}{(A(y) \lif \eq[y][x])}}\nonumber\]
Desarrollamos la derivación hacia atrás:
Ahora tendremos que usar el supuesto principal: al tratarse de una fórmula existencial, utilizamos\(\Elim{\lexists{}{}}\) para derivar la conclusión intermediaria\(\eq[a][b]\).
La subderivación en la parte superior derecha se completa utilizando sus supuestos para mostrar eso\(\eq[a][c]\) y\(\eq[b][c]\). Esto requiere dos derivaciones separadas. La derivación para\(\eq[a][c]\) es la siguiente:
De\(\eq[a][c]\) y\(\eq[b][c]\) derivamos\(\eq[a][b]\) por\(\Elim{\eq[][]}\).
Problema\(\PageIndex{2}\)
Dar derivaciones de las siguientes fórmulas:
-
\(\lforall{x}{\lforall{y}{((\eq[x][y] \land A(x)) \lif A(y))}}\)
-
\(\lexists{x}{A(x)} \land \lforall{y}{\lforall{z}{((A(y) \land A(z)) \lif \eq[y][z])}} \lif \lexists{x}{(A(x) \land \lforall{y}{(A(y) \lif \eq[y][x])})}\)