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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.

Proposición$$\PageIndex{1}$$

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. ◻

Proposición$$\PageIndex{2}$$

$$\Gamma \Proves A$$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$$\Gamma \Sequent A$$:

Problema$$\PageIndex{1}$$

Demostrar que$$\Gamma \Proves \lnot A$$ iff$$\Gamma \cup \{A\}$$ es inconsistente.

Proposición$$\PageIndex{3}$$

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. ◻

Proposición$$\PageIndex{4}$$

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. ◻

This page titled 8.9: Derivabilidad y consistencia is shared under a CC BY license and was authored, remixed, and/or curated by Richard Zach et al. (Open Logic Project) .