Transformation of a Conditional Structure

This chapter focuses on the function $tr_{CS}$ , which was introduced in section 3.1 and allows the transformation of a imbalanced CS into a balanced CS without having further knowledge about the underlaying knowledge base the CS was derived from. Already the results from section 3.2 and section 3.3 show that it is unlikely that $tr_{CS}$ could exist for the case of imbalanced usage of ground atoms. Nevertheless we saw in chapter 10 that for atomic conditionals it seems possible to find out from the CS whether or not the related knowledge base is balanced with respect to the sharing of ground atoms. In this chapter we therefore focus only on knowledge bases which are already balanced with respect to the usage of ground atoms. Based on an example we will see that the transformation of a CS is indeed possible, but only in very limited cases, i.e. the mechanism described here will only work for c-conditionals.

The results shown here are most likely not applicable to real-life cases, but might nevertheless serve as a base for further investigations.

11.1 Initial Definitions and Basic Example

We start with applying the characteristic of balanced and imbalanced not only to knowledge bases but also to CSs.

Definition 49 (Balanced and Imbalanced Conditional Structure)

Let $\kb$ be an FOPCL knowledge base and let $\gamma(\kb)$ be the conditional structure of $\kb$ .

Then $\gamma(\kb)$ is balanced (or imbalanced) iff $\kb$ is balanced (or imbalanced respectively).

More specifically,

there is an balanced sharing (or imbalanced sharing) in $\gamma(\kb)$ iff there is a balanced sharing (or imbalanced sharing respectively) in $\kb$ ,
there is an balanced use (or imbalanced use) in $\gamma(\kb)$ iff there is a balanced use (or imbalanced use respectively) in $\kb$ .

It is $\isBS(\gamma(\kb))$ the function which indicates, whether $\gamma(\kb)$ has a balanced sharing.

It is $\isBU(\gamma(\kb))$ the function which indicates, whether $\gamma(\kb)$ has a balanced use.

The following example 70 will be used as a base throughout this chapter. It makes use of an atomic knowledge base which consists of two c-conditionals which include only local instantiation restrictions. This example has been kept intentionally at a very minimum level, i.e. it only discusses c-conditionals and not cac-conditionals and it also keeps the related transformations at a minimum.

We know from proposition 2 that it is not possible to determine from the CS of a knowledge base whether there’s an imbalanced use in that knowledge base. We therefore restrict ourselves in this chapter to knowledge bases for which it holds that they are balanced with respect to the usage of ground atoms. We will only look into the transformation of CSs which include an imbalanced sharing into their balanced version.

Example 70 (Transformation of Conditional Structure - Basic Example)

Let $\kba_0 = \{~\Ra_1,~\Ra_2~\}$ be an atomic knowledge base with

$\Ra_1 = \big<~\big(~C(V_1)|~\top~\big),~\emptyset~\big>$ and
$\Ra_2 = \big<~\big(~C(V_2)|~\top~\big),~V_2 \neq d~\big>$ ,

with variables $V_1, V_2$ ranging over sort $s_1 = \{~a,~b,~c,~d~\}$ .

It is

$\at(\Ra_1) = \{~C(a),~C(b),~C(c),~C(d)~\}$ and
$\at(\Ra_2) = \{~C(a),~C(b),~C(c)~\}$ .

It holds that $\isBU(\Ra_1) = \true$ and $\isBU(\Ra_2) = \true$ , as both conditionals have only one predicate symbol in the conclusion and none in the antecedence.

It holds that $\isBS(\Ra_1,\Ra_2) = \false$ as $\at(\Ra_1) \cap \at(\Ra_2) \neq \emptyset$ and $\at(\Ra_1) \neq \at(\Ra_2)$ .

Table 11.1 shows the common antecedence table $\Ra_1$ and $\Ra_2$ . It indicates the groundings of $\Ra_1$ in the column " $\top_1$ and the groundings of $\Ra_2$ in the column " $\top_2$ .

Table 11.1
	$\top_1$	$\top_2$
$C(a)$	$r_1$	$r_5$	$\coan_1$
$C(b)$	$r_2$	$r_6$
$C(c)$	$r_3$	$r_7$
$C(d)$	$r_4$		$\coan_2$

Table 11.1: Common Antecedence Sets of Example 70

We now derive the combined vf truth table of $\Rc_1$ and $\Rc_2$ , which is shown in table 11.2. Table 11.2 already indicates also all those vf-pairs which are generated by further conditionals used in this chapter. This is done in order to reduce the number of tables.

Table 11.2
	$\Theta_1$			$\Theta_2$	$\Rc_1$	$\Rc_2$	$\Rc_{1-1}$	$\Rc_{1-2}$
	$C(a)$	$C(b)$	$C(c)$	$C(d)$	$\top_1$	$\top_2$	$\top_{1-1}$	$\top_{1-2}$
$\Theta_1\timesM\Theta_2$					$\omega^{a_1}_{(\top)}$	$\omega^{a_2}_{(\top)}$	$\omega^{a_{1-1}}_{(\top)}$	$\omega^{a_{1-2}}_{(\top)}$
$\theta^{\|0\|}_1 \timesM \theta^{\|0\|}_2$	$0$	$0$	$0$	$0$	$(0,4)$	$(0,3)$	$(0,1)$	$(0,3)$
$\theta^{\|0\|}_1 \timesM \theta^{\|1\|}_2$	$0$	$0$	$0$	$1$	$(1,3)$	$(0,3)$	$(1,0)$	$(0,3)$
$\theta^{\|1\|}_1 \timesM \theta^{\|0\|}_2$	$0$	$0$	$1$	$0$	$(1,3)$	$(1,2)$	$(0,1)$	$(1,2)$
$\theta^{\|1\|}_1 \timesM \theta^{\|1\|}_2$	$0$	$0$	$1$	$1$	$(2,2)$	$(1,2)$	$(1,0)$	$(1,2)$
$\theta^{\|2\|}_1 \timesM \theta^{\|0\|}_2$	$0$	$1$	$1$	$0$	$(2,2)$	$(2,1)$	$(0,1)$	$(2,1)$
$\theta^{\|2\|}_1 \timesM \theta^{\|1\|}_2$	$0$	$1$	$1$	$1$	$(3,1)$	$(2,1)$	$(1,0)$	$(2,1)$
$\theta^{\|3\|}_1 \timesM \theta^{\|0\|}_2$	$1$	$1$	$1$	$0$	$(3,1)$	$(3,0)$	$(0,1)$	$(3,0)$
$\theta^{\|3\|}_1 \timesM \theta^{\|1\|}_2$	$1$	$1$	$1$	$1$	$(4,0)$	$(3,0)$	$(1,0)$	$(3,0)$
					$[4]$	$[3]$	$[1]$	$[3]$

Table 11.2: Combined vf Truth Table of Example 70

We can read the CS of $\kba_0$ directly from table 11.1, i.e. it is

$\gamma(\kba_0) = \{$	$(~(0,4),~(0,3)~),$
	$(~(1,3),~(0,3)~),$	$(~(1,3),~(1,2)~),$
	$(~(2,2),~(1,2)~),$	$(~(2,2),~(2,1)~),$
	$(~(3,1),~(2,1)~),$	$(~(3,1),~(3,0)~),$
	$(~(4,0),~(3,0)~)$		$\}$

From this we can derive the CA-table of $\Rc_1$ and $\Rc_2$ as shown in table 11.3.

Table 11.3
		$[3]$
$CA(\Rc_1,\Rc_2)$		$(0,3)$	$(1,2)$	$(2,1)$	$(3,0)$
$[4]$	$(0,4)$	$*$
	$(1,3)$	$*$	$*$
	$(2,2)$		$*$	$*$
	$(3,1)$			$*$	$*$
	$(4,0)$				$*$

Table 11.3: CA-Table of

R^c_1

and

R^c_2

of Example 70

With the findings of chapter 10 it follows that $\gamma(\kba)$ has an imbalanced sharing as the CA-table is neither completely filled nor does it include a CA-block with only a single CA-line.

The conditional contribution of $\Rc_1$ is $\gamma(\Rc_1) = \{~[4]~\}$ and the conditional contribution of $\Rc_2$ is $\gamma(\Rc_2) = \{~[3]~\}$ .

11.2 Transformation

Due to the imbalanced sharing between $\Rc_1$ and $\Rc_2$ , $\kba_0$ is not parametrically uniform. We will now apply the necessary transformation from section 2.2 and [2] to transform $\kba_0$ into a parametrically uniform knowledge base $\PU(\kba_0)$ .

Example 71 (Transformation)

Continuing from example 70.

We apply transformation rule $TE_1$ to $\Rc_1$ of $\kba_0$ and get the transformed knowledge base $\kba_1 = \{~\Rc_{1-1},~\Rc_{1-2}~\Rc_2~\}$ with

$\Rc_{1-1} = \big<~\big(~C(V_1)|~\top~\big),~V_1=d~\big>$ ,
$\Rc_{1-2} = \big<~\big(~C(V_1)|~\top~\big),~V_1\neq d~\big>$ and
$\Rc_2 = \big<~\big(~C(V_2)|~\top~\big),~V_2\neq d~\big>$ .

We first look at the common antecedence table of the transformed knowledge base as shown in table 11.4.

Table 11.4
	$\top_{1-1}$	$\top_{1-2}$	$\top_2$
$C(a)$		$r_1$	$r_5$	$\coan_1$
$C(b)$		$r_2$	$r_6$
$C(c)$		$r_3$	$r_7$
$C(d)$	$r_4$			$\coan_2$

Table 11.4: Common Antecedence Table of Example 71

As the combined common antecedence sets are the same as in example 70, we can read the CS of $\kba_1$ from the combined vf truth table which is shown in table 11.2

$\gamma(\kba) = \{$	$(~(0,1),~(0,3),~(0,3)~),$	$(~(1,0),~(0,3),~(0,3)~),$
	$(~(0,1),~(1,2),~(1,2)~),$	$(~(1,0),~(1,2),~(1,2)~),$
	$(~(0,1),~(2,1),~(2,1)~),$	$(~(1,0),~(2,1),~(2,1)~),$
	$(~(0,1),~(3,0),~(3,0)~),$	$(~(1,0),~(3,0),~(3,0)~),$	$\}$

It is obvious that $\Ra_1$ and $\Ra_{1-2}$ are identical and therefore are balanced with respect to sharing. We therefore only look into the CA-table of $\Ra_{1-1}$ and $\Ra_{2}$ , which is shown in table 11.5.

Table 11.5
		$[2]$
$CA(\Rc_{1-1},\Rc_{1-2 })$		$(0,2)$	$(1,1)$	$(2,0)$
$[1]$	$(0,1)$	$*$	$*$	$*$
$[1]$	$(1,0)$	$*$	$*$	$*$

Table 11.5: CA-Table of

R^c_{1-1}

and

R^c_{1-2}

of Example 71

The CA-table is full and therefore the two conditionals are balanced with respect to sharing.

The conditional contribution of $\Rca_{1-1}$ is $\gamma(\Rca_{1-1}) = \{~[1]~\}$ and the conditional contribution of $\Rca_{1-2}$ is $\gamma(\Rca_{1-2}) = \{~[1]~\}$ .

We saw in example 70 that the conditional contribution of $\Rc_1$ is $\gamma(\Rc_1) = \{~[4]~\}$ and now, after the transformation, we see in example 71 that $\gamma(\Rca_{1-1}) = \{~[1]~\}$ and $\gamma(\Rca_{1-2}) = \{~[3]~\}$ . This is obvious, as we changed the number of counted c-atoms from 3 (for $\Rc_1$ ) to 1 (for $\Rc_{1-1})$ and 2 (for $\Rc_{1-2}$ ). The transformation cannot change the sum of counted c-atoms, else the semantical content reflected in the knowledge base would be changed. This is a general rule that applies to all transformations which only split the number of c-atoms assigned to a c-conditional to two (transformed) c-conditionals. Therefore the following assumption seems to be correct.

Assumption 7 (Transforming CS of C-Conditionals).

Let $\Rc_1$ and $\Rc_2$ be be two c-conditionals for which it holds that $\isBU(\Rc_1)$ and $\isBU(\Rc_2)$ and $\neg\isBS(\Rc_1,\Rc_2)$ . Let $\Rc_{1-1}$ and $\Rc_{1-2}$ be the c-conditionals which result from the transformation of $\Rc_{1}$ , so that the resulting knowledge base $\kba=\{~\Rc_{1-1},~\Rc_{1-2},~\Rc_{2}~\}$ is balanced.

Let $[n_1]$ be the MOS of $\Rc_{1}$ , $[n_{1-1}]$ and $[n_{1-2}]$ the MOS of $\Rc_{1-1}$ and $\Rc_{1-2}$ respectively.

Then it holds

that $n_1 = n_{1-1} + n_{1-2}$ ; and
that either $n_{1-1} = |\ovl(\Rc_1,\Rc_2)|$ and $n_{1-2} = |\diff(\Rc_1,\Rc_2)|$ or vice versa.

11.3 Discussion

Assumption 7 only is valid for very restricted scenarios, i.e. for the case of two c-conditionals which are both are balanced with respect to the usage of ground atoms, which hold a imbalanced sharing amongst each other and for which a single transformation is enough to make them balanced. It does not hold for the more complex case of ca-conditionals or cc-conditionals, as for those also the amount of a-atoms can vary.

As we have seen in chapter 5 and chapter 6, the contributions of composed a-segments are the sums of several canonical a-segments. In the case of calculation in chapter 6 we constructed the CS of two ca-conditionals by means of multiplications between the numbers of a-atoms and c-atoms. This resulted in MOSs which are the result of multiplications and from the pure number of the MOS and the vf-pairs it is therefore not anymore possible to read which numbers were multiplied.

This means that we would need to have further knowledge about the structure of the given knowledge base, specifically about the distribution of c-atoms and a-atoms, in order to transform the CS of ca-conditionals into a parametrically uniform version. But such knowledge would contradict the aim of this chapter, i.e. to perform transformations solely based on CSs. Additionally, we have seen in section 6.5 that the arithmetic of MOS is not straight forward, therefore it cannot be said how MOSs should be summed up or multiplied with each other in the cases of cac-conditionals.

Still it is interesting to see that the MOS, which we initially only introduced to make notation of conditional contributions easier, seems to play a significant role when it comes to the transformation of atomic knowledge bases.