Mathematische Logik/Gemischte Definitionsabfrage/11/Aufgabe/Lösung

${\displaystyle {}\alpha }$

${\displaystyle {}\Gamma }$

${\displaystyle {}\alpha _{1},\ldots ,\alpha _{n}\in \Gamma }$

${\displaystyle \vdash \alpha _{1}\wedge \ldots \wedge \alpha _{n}\rightarrow \alpha }$

gilt.

${\displaystyle {}\preccurlyeq }$

1. Es ist ${\displaystyle {}i\preccurlyeq i}$ für alle ${\displaystyle {}i\in I}$.
2. Aus ${\displaystyle {}i\preccurlyeq j}$ und ${\displaystyle {}j\preccurlyeq k}$ folgt stets ${\displaystyle {}i\preccurlyeq k}$.
3. Aus ${\displaystyle {}i\preccurlyeq j}$ und ${\displaystyle {}j\preccurlyeq i}$ folgt ${\displaystyle {}i=j}$.

${\displaystyle {}m,n\in M}$

${\displaystyle {}\alpha \in L_{1}^{S}}$

${\displaystyle {}x}$

${\displaystyle {}\lambda }$

${\displaystyle {}M}$

${\displaystyle I{\frac {m}{x}}\vDash \alpha {\text{ genau dann, wenn }}I{\frac {n}{x}}\vDash \alpha }$

gilt.

${\displaystyle {}\mu }$

${\displaystyle (M,R,\mu )\vDash \alpha }$

gilt.