More relations for infons

Now, that we understand what a state of affairs, an infon and what a situoid is, we will mention how an infon behaves, how it can enter into relations. There are relations between situoids or situation and infons, and relations between infons.

We will consider a relation $ \Rightarrow_l \subseteq Infon \times
Infon$. We want to understand $\phi \Rightarrow_l \phi'$ as ``$\phi$ is as strong as $\phi'$''. This relation is reflexive and transitive, and if $\phi \Rightarrow_l \phi'$ and $\phi$ are facts of a situoid, so is $\phi'$. We can capture this in the following axioms.

Axiom 5.14  

$\displaystyle \forall \phi (\phi \Rightarrow_l \phi)$

$\displaystyle \forall \phi \phi' \phi'' (\phi \Rightarrow_l \phi' \land \phi' \Rightarrow_l
\phi'' \rightarrow \phi \Rightarrow_l \phi'')
$

$\displaystyle \forall s \forall \phi \forall \phi' ( s \models \langle\langle \...
..., \phi,
\phi' \rangle\rangle \land s \models \phi \rightarrow s \models \phi')
$

This axiom will be needed when we write more about parameters of infons and the anchor function.

Barwise introduced another operation on infons, allowing us to merge or unify compatible infons. What does it mean for two infons to be compatible?

Definition 5.13 (Compatibility of infons)   Two infons $\phi=\langle\langle R,f;i\rangle\rangle $ and $ \phi'=\langle\langle R',f',i'\rangle\rangle $ are compatible, if and only if $R=R'$, $i=i'$ and $f$ and $ f'$ are compatible as functions.

Now we can state what is meant by merging two compatible infons.

Axiom 5.15   If $\phi=\langle\langle R,f;i\rangle\rangle $ and $\phi'=\langle\langle R,f';i\rangle\rangle $ are compatible infons, than

$\displaystyle \phi \oplus \phi' = \langle\langle R,f \cup f';i\rangle\rangle
$

This function is mainly used to fill missing arguments of infons, and we will not use it further.

At last, we will assert that every infon has a complement.

Axiom 5.16   For all infons $ \phi=\langle\langle R,a;p \rangle\rangle $ exists another infon denoted as $ \bar{\phi}$ with $ \bar{\phi} = \langle\langle R,a;1-p \rangle\rangle $.

leechuck 2005-04-19