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Homeproof that $C_\cup$ and $C_\cap$ are consequence operators

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# proof that $C_{\cup}$ and $C_{\cap}$ are consequence operators

The proof that the operators $C_{\cup}$ and $C_{\cap}$ defined in the second example of section 3 of the parent entry are consequence operators is a relatively straightforward matter of checking that they satisfy the defining properties given there. For convenience, those definitions are reproduced here.

###### Definition 1.

Given a set $L$ and two elements, $X$ and $Y$, of this set, the function $C_{\cap}(X,Y)\colon\mathcal{P}(L)\to\mathcal{P}(L)$ is defined as follows:

$C_{\cap}(X,Y)(Z)=\begin{cases}X\cup Z&Y\cap Z\not=\emptyset\\ Z&Y\cap Z=\emptyset\end{cases}$ |

###### Theorem 1.

For every choice of two elements, $X$ and $Y$, of a given set $L$, the function $C_{\cap}(X,Y)$ is a consequence operator.

###### Proof.

*Property 1:*
Since $Z$ is a subset of itself and of $X\cup Z$, it follows that
$Z\subseteq C_{\cap}(X,Y)(Z)$ in either case.

*Property 2:*
We consider two cases. If $Y\cap Z=\emptyset$, then $C_{\cap}(X,Y)(Z)=Z$, so

$C_{\cap}(X,Y)(C_{\cap}(X,Y)(Z))=C_{\cap}(X,Y)(Z).$ |

If $Y\cap Z\not=\emptyset$, then

$\displaystyle Y\cap C_{\cap}(X,Y)(Z)$ | $\displaystyle=$ | $\displaystyle Y\cap(X\cup Z)$ | ||

$\displaystyle=$ | $\displaystyle(Y\cap X)\cup(Y\cap Z).$ |

Again, since $Y\cap Z\not=\emptyset$, we also have $(Y\cap X)\cup(Y\cap Z)\not=\emptyset$, so

$\displaystyle C_{\cap}(X,Y)(C_{\cap}(X,Y)(Z))$ | $\displaystyle=$ | $\displaystyle X\cup C_{\cap}(X,Y)(Z)$ | ||

$\displaystyle=$ | $\displaystyle X\cup(X\cup Z)$ | |||

$\displaystyle=$ | $\displaystyle X\cup Z$ | |||

$\displaystyle=$ | $\displaystyle C_{\cap}(X,Y)(Z)$ |

So, in both cases, we find that

$C_{\cap}(X,Y)(C_{\cap}(X,Y)(Z))=C_{\cap}(X,Y)(Z).$ |

*Property 3:*
Suppose that $Z$ and $W$ are subsets of $L$ and that $Z$ is a subset
of $W$. Then there are three possibilities:

1. $Y\cap Z=\emptyset$ and $Y\cap W=\emptyset$

In this case, we have $C_{\cap}(X,Y)(Z)=Z$ and $C_{\cap}(X,Y)(W)=W$, so $C_{\cap}(X,Y)(Z)\subseteq C_{\cap}(X,Y)(W)$.

2. $Y\cap Z=\emptyset$ but $Y\cap W\not=\emptyset$

In this case, $C_{\cap}(X,Y)(Z)=Z$ and $C_{\cap}(X,Y)(W)=X\cup W$. Since $Z\subseteq W$ implies $Z\subseteq X\cup W$, we have $C_{\cap}(X,Y)(Z)\subseteq C_{\cap}(X,Y)(W)$.

3. $Y\cap Z\not=\emptyset$ and $Y\cap W\not=\emptyset$

In this case, $C_{\cap}(X,Y)(Z)=X\cup Z$ and $C_{\cap}(X,Y)(W)=X\cup W$. Since $Z\subseteq W$ implies $X\cup Z\subseteq X\cup W$, we have $C_{\cap}(X,Y)(Z)\subseteq C_{\cap}(X,Y)(W)$.

∎

###### Definition 2.

Given a set $L$ and two elements, $X$ and $Y$, of this set, the function $C_{\cup}(X,Y)\colon\mathcal{P}(L)\to\mathcal{P}(L)$ is defined as follows:

$C_{\cup}(X,Y)(Z)=\begin{cases}X\cup Z&Y\cup Z=Z\\ Z&Y\cup Z\not=Z\end{cases}$ |

###### Theorem 2.

For every choice of two elements, $X$ and $Y$, of a given set $L$, the function $C_{\cup}(X,Y)$ is a consequence operator.

###### Proof.

*Property 1:*
Since $Z$ is a subset of itself and of $X\cup Z$, it follows that
$Z\subseteq C_{\cup}(X,Y)(Z)$ in either case.

*Property 2:*
We consider two cases. If $C_{\cup}(X,Y)(Z)=Z$, then

$C_{\cup}(X,Y)(C_{\cup}(X,Y)(Z))=C_{\cup}(X,Y)(Z).$ |

If $C_{\cup}(X,Y)(Z)=X\cup Z$, then we note that, because $X\cup(X\cup Z)=X\cup Z$, we must have $C_{\cup}(X,Y)(X\cup Z)=X\cup Z$ whether or not $Y\cup(X\cup Z)=X\cup Z$, so

$C_{\cup}(X,Y)(C_{\cup}(X,Y)(Z))=C_{\cup}(X,Y)(Z).$ |

*Property 3:*
Suppose that $Z$ and $W$ are subsets of $L$ and that $Z$ is a subset
of $W$. Then there are three possibilities:

1. $Y\cup Z=Z$ and $Y\cup W=W$

In this case, we have $C_{\cup}(X,Y)(Z)=X\cup Z$ and $C_{\cup}(X,Y)(W)=X\cup W$. Since $Z\subseteq W$ implies $X\cup Z\subseteq X\cup W$, we have $C_{\cup}(X,Y)(Z)\subseteq C_{\cup}(X,Y)(W)$.

2. $Y\cup Z\not=Z$ but $Y\cup W=W$

In this case, $C_{\cup}(X,Y)(Z)=Z$ and $C_{\cup}(X,Y)(W)=X\cup W$. Since $Z\subseteq W$ implies $Z\subseteq X\cup W$, we have $C_{\cup}(X,Y)(Z)\subseteq C_{\cup}(X,Y)(W)$.

3. $Y\cup Z\not=Z$ and $Y\cup W\not=W$

In this case, $C_{\cup}(X,Y)(Z)=Z$ and $C_{\cup}(X,Y)(W)=W$, so $C_{\cup}(X,Y)(Z)\subseteq C_{\cup}(X,Y)(W)$. ∎

## Mathematics Subject Classification

03G25*no label found*03G10

*no label found*03B22

*no label found*

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## Comments

## preamble problem

When I last had a look at the entry two weeks ago, it was fine.

Now, when I revisited it, it was broken. The computer had thrown

away the preamble and replaced it with the default preamble (in

which "theorem", "definition", and "proof" are undefined (although

they really should be)). Has anyone else had this experience?

This, alomg with the problem of the entry which was shown to be

an attachement when it was not, suggests that there might be some

instability in the database.

## Re: preamble problem

I've never had this problem. I suspect it's rather rare: I've just Googled for broken entries on PlanetMath, and yours (now fixed) is the only one that comes up.

I think the attachment problem was caused by the parent being deleted. (If there were a log of deletions, as has been suggested before, this would be easy to check.) The fact that the "Attached to" box appears empty doesn't surprise me, as I've noticed before that this information doesn't appear to be stored as text (since the canonical names I enter are often changed to non-canonical names when I next edit the entry) - the box may be empty not because there is no parent specified, but because the system cannot find a name for the specified parent (because it's been deleted).