| Version 4 |
Version 3 |
| The proof that the operators $C_\cup$ and $C_\cap$ defined in the second |
The proof that the operators $C_\cup$ and $C_\cap$ defined in the second |
| example of section 3 of the |
example of section 3 of the |
| \PMlinkname{parent entry}{ConsequenceOperator} are consequence operators |
\PMlinkname{parent entry}{ConsequenceOperator} are consequence operators |
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is a relatively straightforward matter of checking that they satisfy the defining properties given there. Recall that these operators are defined
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is a relatively straightforward matter of checking that they satisfy the defining |
| as follows: |
properties given there. |
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| \begin{definition} |
\begin{theorem} $C_\cup$ is a consequence operator. \end{theorem} |
| \begin{eqnarray*} |
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| C_\cap (X,Y)(Z) &=& \begin{cases} X \cup Z & Y \cap Z |
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| \not= \emptyset \\ Z & Y \cap Z = \emptyset \end{cases} \\ |
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| C_\cup (X,Y)(Z) &=& \begin{cases} X \cup Z & Y \cup Z = Z \\ Z & Y \cup Z |
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| \not= Z \end{cases} |
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| \end{eqnarray*} |
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| \end{definition} |
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| \begin{theorem} $C_\cap$ is a consequence operator. \end{theorem} |
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\begin{proof} ~
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\begin{proof} |
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| \emph{Property 1:} |
\emph{Property 1:} |
| Since $Z$ is a subset of itself and of $Z \cap X$, it follows that |
Since $Z$ is a subset of itself and of $Z \cap X$, it follows that |
| $Z \subseteq C_\cup (X,Y) (Z)$ in either case. |
$Z \subseteq C_\cup (X,Y) (Z)$ in either case. |
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| \emph{Property 2:} |
\emph{Property 2:} |
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| \emph{Property 3:} |
\emph{Property 3:} |
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| \end{proof} |
\end{proof} |
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\begin{theorem} $C_\cup$ is a consequence operator. \end{theorem}
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\begin{theorem} $C_\cap$ is a consequence operator. \end{theorem}
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\begin{proof} ~
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\begin{proof} |
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| \emph{Property 1:} |
\emph{Property 1:} |
| Since $Z$ is a subset of itself and of $Z \cap X$, it follows that |
Since $Z$ is a subset of itself and of $Z \cap X$, it follows that |
| $Z \subseteq C_\cup (X,Y) (Z)$ in either case. |
$Z \subseteq C_\cup (X,Y) (Z)$ in either case. |
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| \emph{Property 2:} |
\emph{Property 2:} |
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| \emph{Property 3:} |
\emph{Property 3:} |
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| \end{proof} |
\end{proof} |
