Qualitative trust model with a configurable method to aggregate ordinal data

Abstract

Trust models are mechanisms that allow agents to build trust without relying on a trusted central authority. Our goal was to develop a trust model that would operate with values that humans easily understand and manipulate: qualitative and ordinal values. The result is a trust model that computes trust from experiences created in interactions and from opinions obtained from third-party agents. The trust model, termed qualitative trust model (QTM), uses qualitative and ordinal values for assessing experiences, expressing opinions and estimating trust. We treat such values appropriately; we never convert them to numbers, but merely use their relative order. To aggregate a collection of such values, we propose an aggregation method that is based on comparing distributions and show some of its properties; the method can be used in other domains and can be seen as an alternative to median and similar methods. To cope with lying agents, QTM estimates trustworthiness in opinion providers with a modified version of the weighted majority algorithm, and additionally combines trustworthiness with social links between agents; such links are obtained implicitly by observing how agents provide opinions about each other. Finally, we compare QTM against a set of well-known trust models and demonstrate that it consistently performs well and on par with other quantitative models, and in many cases even outperforms them, particularly when the number of direct experiences is low.

Appendix

We first introduce terms that we use in the construction.

Lemma 1

Let $\mathbf {p}=[f_{D}, f_{\textit{PD}}, f_{I}, f_{PT}, f_{T}]$ be a distribution vector. Then

• (a) $0 \le f_\omega \le 1$ for all $\omega \in \Omega$,

• (b) $\sum _{\omega \in \Omega } f_\omega = 1$.

Proof

Both properties follow directly from Definition 2. $\square$

Since the distance metric in Eq. (1) includes a nested expression, which is cumbersome to manipulate, we introduce the notion of cumulative distribution vectors.

Definition 6

Let $\mathbf {p}=[f_{D}, f_{\textit{PD}}, f_{I}, f_{\textit{PT}}, f_{T}]$ be a distribution vector. Let a cumulative distribution vector $\mathbf {P} = [F_D, F_{\textit{PD}}, F_I, F_{\textit{PT}}, F_T]$ be computed by successively summing up components of $\mathbf {p}$ as follows.

$$\begin{aligned} F_\omega = \sum _{\omega _x \in \Omega }^{\omega _x \le \omega } f_{\omega _x} \end{aligned}$$

We denote distribution vectors and their components with a lower-case letter, such as $\mathbf {p}$ and $f_\textit{PT}$, while we write cumulative distribution vectors and their components with upper-case letters, for instance $\mathbf {P}$ and $F_\textit{PT}$.

Lemma 2

Let $\mathbf {P} = [F_D, F_{\textit{PD}}, F_I, F_{PT}, F_T]$ be a cumulative distribution vector. Then

• (a) $F_D \le F_{\textit{PD}} \le F_I \le F_{PT} \le F_T$,

• (b) $0 \le F_D$,

• (c) $F_T = 1$.

Proof

All properties follow directly from previously defined terms.

• (a) Follows directly from Definition 6.

• (b) Follows directly from Definition 6 and Lemma 1.

• (c) Follows directly from Definition 6 and Lemma 1. $\square$

We rewrite the metric from Eq. (1) to accommodate cumulative distribution vectors.

$$\begin{aligned} D(\mathbf {X}, \mathbf {Y}) = \sqrt{ \sum _{\omega \in \Omega } \left( \mathbf {X}(\omega ) - \mathbf {Y}(\omega ) \right) ^2} \end{aligned}$$

(2)

Lemma 3

Let $\mathbf {x}$ and $\mathbf {y}$ be two distribution vectors, and let $\mathbf {X}$ and $\mathbf {Y}$ their cumulative versions as described in Definition 6. Then

$$\begin{aligned} d(\mathbf {x}, \mathbf {y}) = D(\mathbf {X}, \mathbf {Y}) \end{aligned}$$

Proof

The equality follows directly from definitions of metrics in Eqs. (1) and (2) and the definition of cumulative distribution vectors in Definition 6. $\square$

1.1 Equality with median

First, we define the median of a collection. “The median of a collection of values can be found by arranging all the observations from lowest value to highest value and picking the middle one. If there is an even number of observations, then there is no single middle value; the median is then usually defined to be the mean of the two middle values” [42]. But since we operate with qualitative values, we cannot compute the mean. In such cases we pick the value that is the highest between the two. In canonical form, we define the ${{\mathrm{Median}}}$ of a cumulative distribution vector $\mathbf {P}$ as in Eq. (3).

$$\begin{aligned} {{\mathrm{Median}}}(\mathbf {P}) = \left\{ \begin{array}{l@{\quad }l} D &{} F_D > \frac{1}{2}\\ { PD} &{} F_D \le \frac{1}{2} \wedge F_{\textit{PD}} > \frac{1}{2} \\ I &{} F_\textit{PD} \le \frac{1}{2} \wedge F_I > \frac{1}{2} \\ { PT} &{} F_I \le \frac{1}{2} \wedge F_{PT} > \frac{1}{2}\\ T &{} F_{PT} \le \frac{1}{2} \end{array}\right. \end{aligned}$$

(3)

Second, we compute representative cumulative distribution vectors from representative distribution vectors given in Theorem 1: $\mathbf {R}_{D} = [1,1,1,1,1]$, $\mathbf {R}_{\textit{PD}} = [0,1,1,1,1]$, $\mathbf {R}_{I} = [0,0,1,1,1]$, $\mathbf {R}_{PT} = [0,0,0,1,1]$, $\mathbf {R}_{T} = [0,0,0,0,1]$.

Finally, we proceed with the proof. The construction is straightforward: given a cumulative distribution vector, $\mathbf {P}$, for which the ${{\mathrm{Median}}}$ returns a certain value, $\omega ={{\mathrm{Median}}}(\mathbf {P})$, we show that the cumulative representative distribution vector that belongs to the same value, $\mathbf {R}_\omega$, is the closest to $\mathbf {P}$.

Proof

Let us have a distribution vector $\mathbf {p}=[f_{D}, f_{\textit{PD}}, f_{I}, f_{PT}, f_{T}]$ and its cumulative version $\mathbf {P} = [F_D, F_{\textit{PD}}, F_I, F_{PT}, F_T]$. First, we compute the distance between the cumulative distribution vector $\mathbf {P}$ and all cumulative representative distribution vectors.

$$\begin{aligned} D(\mathbf {P}, \mathbf {R}_{D})&= \sqrt{(1-F_D)^2 + (1-F_{\textit{PD}})^2 + (1-F_I)^2 + (1-F_{PT})^2} \end{aligned}$$

(4)

$$\begin{aligned} D(\mathbf {P}, \mathbf {R}_{PD})&= \sqrt{{F_D}^2 + (1-F_{\textit{PD}})^2 + (1-F_I)^2 + (1-F_{PT})^2} \end{aligned}$$

(5)

$$\begin{aligned} D(\mathbf {P}, \mathbf {R}_{I})&= \sqrt{{F_D}^2 + {F_{\textit{PD}}}^2 + (1-F_I)^2 + (1-F_{PT})^2} \end{aligned}$$

(6)

$$\begin{aligned} D(\mathbf {P}, \mathbf {R}_{PT})&= \sqrt{{F_D}^2 + {F_{\textit{PD}}}^2 + {F_I}^2 + (1-F_{PT})^2} \end{aligned}$$

(7)

$$\begin{aligned} D(\mathbf {P}, \mathbf {R}_{T})&= \sqrt{{F_D}^2 + {F_{\textit{PD}}}^2 + {F_I}^2 + {F_{PT}}^2} \end{aligned}$$

(8)

The first case we check is $D = {{\mathrm{Median}}}(\mathbf {P})$. Here, we have to prove the following.

$$\begin{aligned} \begin{array}{l} F_D > \frac{1}{2} \end{array}\iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_{\textit{PD}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_I)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_{PT})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_T) \end{array} \end{aligned}$$

(9)

We use expressions from Eqs. (4)–(8), apply basic arithmetic operations and the Eq. (9) simplifies to the following.

$$\begin{aligned} \begin{array}{l} F_D > \frac{1}{2} \end{array} \iff \begin{array}{l} \frac{1}{2} < F_D\, \wedge \\ 1 < F_D + F_{\textit{PD}} \,\wedge \\ \frac{3}{2} < F_D + F_{\textit{PD}} + F_I\, \wedge \\ 2 < F_D + F_{\textit{PD}} + F_I + F_{PT} \end{array} \end{aligned}$$

(10)

Because of Lemma 2, the first component on the right side is the strictest, thus we can drop the remaining three. We obtain the desired result.

$$\begin{aligned} F_D > \frac{1}{2} \iff \frac{1}{2} < F_D \end{aligned}$$

(11)

The second case covers $PD = {{\mathrm{Median}}}(\mathbf {P})$. Now, we have to show the following.

$$\begin{aligned} \begin{array}{l} F_D \le \frac{1}{2} \wedge F_PD > \frac{1}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_{\textit{PD}}) \le D(\mathbf {P}, \mathbf {R}_D)\, \wedge \\ D(\mathbf {P}, \mathbf {R}_{\textit{PD}}) < D(\mathbf {P}, \mathbf {R}_I) \,\wedge \\ D(\mathbf {P}, \mathbf {R}_{\textit{PD}}) < D(\mathbf {P}, \mathbf {R}_{PT})\, \wedge \\ D(\mathbf {P}, \mathbf {R}_{\textit{PD}}) < D(\mathbf {P}, \mathbf {R}_T) \end{array} \end{aligned}$$

(12)

We use the greater-or-equal-to operator on the first component on the right side of Eq. (12), because in cases when $D(\mathbf {P}, \mathbf {R}_D) = D(\mathbf {P}, \mathbf {R}_{\textit{PD}})$, the aggregation method selects PD (the highest between $D$ and PD; see Definition 4, where we use the maximal operator). We plug in Eqs. (4)–(8) and the expression simplifies.

$$\begin{aligned} \begin{array}{l} F_D \le \frac{1}{2} \wedge F_{\textit{PD}} > \frac{1}{2} \end{array} \iff \begin{array}{l} \frac{1}{2} \ge F_D\,\wedge \\ \frac{1}{2} < F_{\textit{PD}}\, \wedge \\ 1 < F_{\textit{PD}} + F_I\,\wedge \\ \frac{3}{2} < F_{\textit{PD}} + F_I + F_{PT} \end{array} \end{aligned}$$

(13)

Because of Lemma 2, the second component on the right side is stricter than third and forth, thus we can omit them. We obtain the desired result.

$$\begin{aligned} F_D \le \frac{1}{2} \wedge F_{\textit{PD}} > \frac{1}{2} \iff \frac{1}{2} \ge F_D \wedge \frac{1}{2} < F_{\textit{PD}} \end{aligned}$$

(14)

In the third case, we cover $I = {{\mathrm{Median}}}(\mathbf {P})$. We show that the following holds.

$$\begin{aligned} \begin{array}{l} F_{\textit{PD}} \le \frac{1}{2} \wedge F_I > \frac{1}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_I) \le D(\mathbf {P}, \mathbf {R}_D)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_I) \le D(\mathbf {P}, \mathbf {R}_{\textit{PD}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_I) < D(\mathbf {P}, \mathbf {R}_{PT})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_I) < D(\mathbf {P}, \mathbf {R}_T) \end{array} \end{aligned}$$

(15)

Again, plugging in Eqs. (4)–(8) yields Eq. (16).

$$\begin{aligned} \begin{array}{l} F_{\textit{PD}} \le \frac{1}{2} \wedge F_I > \frac{1}{2} \end{array} \iff \begin{array}{l} 1 \ge F_D + F_{\textit{PD}} \,\wedge \\ \frac{1}{2} \ge F_{\textit{PD}}\, \wedge \\ \frac{1}{2} < F_I \,\wedge \\ 1 < F_I + F_{PT} \end{array} \end{aligned}$$

(16)

Lemma 2 ensures that the second component on the right side of Eq. (16) binds stricter than the first, while the third component binds stricter than the forth. Omitting the first and forth thus leads to the following.

$$\begin{aligned} F_{\textit{PD}} \le \frac{1}{2} \wedge F_I > \frac{1}{2} \iff \frac{1}{2} \ge F_{\textit{PD}} \wedge \frac{1}{2} < F_I \end{aligned}$$

(17)

Now, we check $PT = {{\mathrm{Median}}}(\mathbf {P})$. We show the following.

$$\begin{aligned} \begin{array}{l} F_I \le \frac{1}{2} \wedge F_{PT} > \frac{1}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_{PT}) \le D(\mathbf {P}, \mathbf {R}_D)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{PT}) \le D(\mathbf {P}, \mathbf {R}_{\textit{PD}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{PT}) \le D(\mathbf {P}, \mathbf {R}_I)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{PT}) < D(\mathbf {P}, \mathbf {R}_T) \end{array} \end{aligned}$$

(18)

Substituting in Eqs. (4)–(8) yields Eq. (19).

$$\begin{aligned} \begin{array}{l} F_I \le \frac{1}{2} \wedge F_{PT} > \frac{1}{2} \end{array} \iff \begin{array}{l} \frac{3}{2} \ge F_D + F_{\textit{PD}} + F_I\,\wedge \\ 1 \ge F_{\textit{PD}} + F_I \,\wedge \\ \frac{1}{2} \ge F_I\, \wedge \\ \frac{1}{2} < F_{PT} \end{array} \end{aligned}$$

(19)

Using Lemma 2 we simplify Eq. (19).

$$\begin{aligned} F_I \le \frac{1}{2} \wedge F_{PT} > \frac{1}{2} \iff \frac{1}{2} \ge F_I \wedge \frac{1}{2} < F_{PT} \end{aligned}$$

(20)

Finally, we verify $T = {{\mathrm{Median}}}(\mathbf {P})$ by proving the following.

$$\begin{aligned} \begin{array}{l} F_{PT} \le \frac{1}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_D)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_{\textit{PD}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_I)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_{PT}) \end{array} \end{aligned}$$

(21)

Using Eqs. (4)–(8) we obtain Eq. (22).

$$\begin{aligned} \begin{array}{l} F_{PT} \le \frac{1}{2} \end{array} \iff \begin{array}{l} 2 \ge F_D + F_{\textit{PD}} + F_I + F_{PT}\,\wedge \\ \frac{3}{2} \ge F_{\textit{PD}} + F_I + F_{PT}\,\wedge \\ 1 \ge F_I + F_{PT}\,\wedge \\ \frac{1}{2} \ge F_{PT} \end{array} \end{aligned}$$

(22)

Because of Lemma 2 we can omit the first three components on the right side, which yields Eq. (23).

$$\begin{aligned} F_{PT} \le \frac{1}{2} \iff \frac{1}{2} \ge F_{PT} \end{aligned}$$

(23)

We have shown that equivalences in Eqs. (9), (12), (15), (18), (21) hold. This thus proves the equality between the ${{\mathrm{Median}}}$ and ${{\mathrm{Aggregate}}}$ method when using representative distribution vectors from Theorem 1. $\square$

1.2 Equality with mean-of-ranks

With the mean-of-ranks we refer to the procedure that computes an average value from a collection of ordinal values by assigning each ordinal value a number, in this case its corresponding rank, and then computing the mean of the collection with the assigned numbers. We give the formula to compute the mean-of-ranks in Eq. (24); we compute it for a given distribution vector. The square brackets denote the nearest integer function, while ${{\mathrm{\textsc {Rank}}}}^{-1}$ stands for inverse mapping from ranks into qualitative values; the mappings are given in Definition 1.

$$\begin{aligned} {{\mathrm{mor}}}(\mathbf {p}) = {{\mathrm{\textsc {Rank}}}}^{-1}[1 \cdot f_D + 2 \cdot f_{\textit{PD}} + 3 \cdot f_I + 4 \cdot f_{PT} + 5 \cdot f_T] \end{aligned}$$

(24)

Analogously, we write the mean-of-ranks of a cumulative distribution vector.

$$\begin{aligned} {{\mathrm{MOR}}}(\mathbf {P}) = {{\mathrm{\textsc {Rank}}}}^{-1}[5 - F_D - F_{\textit{PD}} - F_I - F_{PT}] \end{aligned}$$

(25)

Lemma 4

Let $\mathbf {p}$ be a distribution vector, and let $\mathbf {P}$ be its cumulative version.

$$\begin{aligned} {{\mathrm{mor}}}(\mathbf {p}) = {{\mathrm{MOR}}}(\mathbf {P}) \end{aligned}$$

Proof

The equality follows directly from definitions of mean-of-ranks in Eqs. (24) and (25) and the definition of cumulative distribution vectors in Definition 6. $\square$

Let us rewrite ${{\mathrm{MOR}}}(\mathbf {P})$ of a cumulative distribution vector $\mathbf {P} = [F_D, F_{\textit{PD}}, F_I, F_{PT}, F_T]$ in canonical form, so it will be easier to analyze.

$$\begin{aligned} {{\mathrm{MOR}}}(\mathbf {P}) = \left\{ \begin{array}{l@{\quad }l} T &{} 5 \ge 5 - F_D - F_{\textit{PD}} - F_I - F_{PT} \ge \frac{9}{2}\\ { PT} &{} \frac{9}{2} > 5 - F_D - F_{\textit{PD}} - F_I - F_{PT} \ge \frac{7}{2}\\ I &{} \frac{7}{2} > 5 - F_D - F_{\textit{PD}} - F_I - F_{PT} \ge \frac{5}{2}\\ { PD} &{} \frac{5}{2} > 5 - F_D - F_{\textit{PD}} - F_I - F_{PT} \ge \frac{3}{2}\\ D &{} \frac{3}{2} > 5 - F_D - F_{\textit{PD}} - F_I - F_{PT} \ge 1 \end{array}\right. \end{aligned}$$

(26)

Next, we compute cumulative representative distribution vectors from representative distribution vectors given in Theorem 2: $\mathbf {R}_{D} = [1,1,1,1,1]$, $\mathbf {R}_{\textit{PD}} = \left[ \frac{3}{4},\frac{3}{4},\frac{3}{4},\frac{3}{4},1\right]$, $\mathbf {R}_{I} = \left[ \frac{1}{2},\frac{1}{2},\frac{1}{2},\frac{1}{2},1\right]$, $\mathbf {R}_{PT} = \left[ \frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{1}{4},1\right]$, $\mathbf {R}_{T} = [0,0,0,0,1]$.

Finally, we are ready to proceed with the proof. The construction is the same as with median: given a cumulative distribution vector, $\mathbf {P}$, for which the ${{\mathrm{MOR}}}$ returns a certain value, $\omega ={{\mathrm{MOR}}}(\mathbf {P})$, we prove that the cumulative representative distribution vector that belongs to that value, $\mathbf {R}_\omega$, is the closest to $\mathbf {P}$.

Proof

Let us have a distribution vector $\mathbf {p}=[f_{D}, f_{\textit{PD}}, f_{I}, f_{PT}, f_{T}]$ and its cumulative version $\mathbf {P} = [F_D, F_{\textit{PD}}, F_I, F_{PT}, F_T]$. First, we compute the distance between the cumulative distribution vector $\mathbf {P}$ and all cumulative representative distribution vectors.

We write down the distances between a cumulative distribution vector $\mathbf {P}$ and a all cumulative representative distribution vectors

$$\begin{aligned}&D(\mathbf {P}, \mathbf {R}_{D}) = \sqrt{(1-F_D)^2 + (1-F_{\textit{PD}})^2 + (1-F_I)^2 + (1-F_{PT})^2} \end{aligned}$$

(27)

$$\begin{aligned}&D(\mathbf {P}, \mathbf {R}_{\textit{PD}}) \!=\! \sqrt{\left( \frac{3}{4}\!-\!F_D\right) ^2 \!+\! \left( \frac{3}{4}\!-\!F_{\textit{PD}}\right) ^2 \!+\! \left( \frac{3}{4}\!-\!F_I\right) ^2 \!+\! \left( \frac{3}{4}\!-\!F_{PT}\right) ^2} \end{aligned}$$

(28)

$$\begin{aligned}&D(\mathbf {P}, \mathbf {R}_{I}) = \sqrt{\left( \frac{1}{2}\!-\!F_D\right) ^2 \!+\! \left( \frac{1}{2}\!-\!F_{\textit{PD}}\right) ^2 \!+\! \left( \frac{1}{2}\!-\!F_I\right) ^2 \!+\! \left( \frac{1}{2}\!-\!F_{PT}\right) ^2} \end{aligned}$$

(29)

$$\begin{aligned}&D(\mathbf {P}, \mathbf {R}_{PT}) = \sqrt{\left( \frac{1}{4}\!-\!F_D\right) ^2 \!+\! \left( \frac{1}{4}\!-\!F_{\textit{PD}}\right) ^2 \!+\! \left( \frac{1}{4}-F_I\right) ^2 \!+\! \left( \frac{1}{4}\!-\!F_{PT}\right) ^2} \end{aligned}$$

(30)

$$\begin{aligned}&D(\mathbf {P}, \mathbf {R}_{T}) = \sqrt{{F_D}^2 + {F_{\textit{PD}}}^2 + {F_I}^2 + {F_{PT}}^2} \end{aligned}$$

(31)

The first case we check is $T = {{\mathrm{MOR}}}(\mathbf {P})$. We show that the following holds true.

$$\begin{aligned} \begin{array}{l} 5 \ge 5 - F_D - F_{\textit{PD}} - F_I - F_{PT} \ge \frac{9}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_D)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_{PD})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_I)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_T) \le D(\mathbf {P}, \mathbf {R}_{PT}) \end{array} \end{aligned}$$

(32)

Using Eqs. (27)–(31) we obtain the following.

$$\begin{aligned} \begin{array}{l} \frac{1}{2} \ge F_D + F_{\textit{PD}} + F_I + F_{PT} \ge 0 \end{array} \iff \begin{array}{l} 2 \ge F_D + F_{\textit{PD}} + F_I + F_{PT}\,\wedge \\ \frac{3}{2} \ge F_D + F_{\textit{PD}} + F_I + F_{PT}\,\wedge \\ 1 \ge F_D + F_{\textit{PD}} + F_I + F_{PT} \wedge \,\\ \frac{1}{2} \ge F_D + F_{\textit{PD}} + F_I + F_{PT} \end{array} \end{aligned}$$

(33)

Because the last expression in Eq. (33) is the strictest, we can omit the remaining three. Moreover, because of Lemma 2, case (b), we can add the lower-bound.

$$\begin{aligned} \begin{array}{l} \frac{1}{2} \ge F_D + F_{\textit{PD}} + F_I + F_{PT} \ge 0 \end{array} \iff \begin{array}{l} \frac{1}{2} \ge F_D + F_{\textit{PD}} + F_I + F_{PT} \ge 0 \end{array} \end{aligned}$$

(34)

Next, we address the case of $PT = {{\mathrm{MOR}}}(\mathbf {P})$.

$$\begin{aligned}&\begin{array}{l} \frac{9}{2} > 5 - F_D - F_{{ PD}} - F_I - F_{{ PT}} \ge \frac{7}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_{{ PT}}) \le D(\mathbf {P}, \mathbf {R}_D)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{{ PT}}) \le D(\mathbf {P}, \mathbf {R}_{{ PD}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{{ PT}}) \le D(\mathbf {P}, \mathbf {R}_I)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{{ PT}}) < D(\mathbf {P}, \mathbf {R}_T) \end{array}\end{aligned}$$

(35)

$$\begin{aligned}&\begin{array}{l} \frac{3}{2} \ge F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{1}{2} \end{array} \iff \begin{array}{l} \frac{5}{2} \ge F_D + F_{{ PD}} + F_I + F_{{ PT}}\,\wedge \\ 2 \ge F_D + F_{{ PD}} + F_I + F_{{ PT}}\,\wedge \\ \frac{3}{2} \ge F_D + F_{{ PD}} + F_I + F_{{ PT}} \wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{1}{2} \end{array}\end{aligned}$$

(36)

$$\begin{aligned}&\begin{array}{l} \frac{3}{2} \ge F_D \!+\! F_{{ PD}} \!+\! F_I \!+\! F_{{ PT}} > \frac{1}{2} \end{array} \iff \begin{array}{l} \frac{3}{2} \ge F_D \!+\! F_{{ PD}} \!+\! F_I \!+\! F_{{ PT}} > \frac{1}{2} \end{array} \end{aligned}$$

(37)

Next, the case of $I = {{\mathrm{MOR}}}(\mathbf {P})$.

$$\begin{aligned}&\begin{array}{l} \frac{7}{2} > 5 - F_D - F_{{ PD}} - F_I - F_{{ PT}} \ge \frac{5}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_I) \le D(\mathbf {P}, \mathbf {R}_D)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_I) \le D(\mathbf {P}, \mathbf {R}_{{ PD}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_I) < D(\mathbf {P}, \mathbf {R}_{{ PT}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_I) < D(\mathbf {P}, \mathbf {R}_T) \end{array}\end{aligned}$$

(38)

$$\begin{aligned}&\begin{array}{l} \frac{5}{2} \ge F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{3}{2} \end{array} \iff \begin{array}{l} 3 \ge F_D + F_{{ PD}} + F_I + F_{{ PT}}\,\wedge \\ \frac{5}{2} \ge F_D + F_{{ PD}} + F_I + F_{{ PT}}\,\wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{3}{2} \wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > 1 \end{array}\end{aligned}$$

(39)

$$\begin{aligned}&\begin{array}{l} \frac{3}{2} \ge F_D \!+\! F_{{ PD}} \!+\! F_I \!+\! F_{{ PT}} > \frac{1}{2} \end{array} \iff \begin{array}{l} \frac{3}{2} \ge F_D \!+\! F_{{ PD}} \!+\! F_I \!+\! F_{{ PT}} > \frac{1}{2} \end{array} \end{aligned}$$

(40)

And the case of ${ PD} = {{\mathrm{MOR}}}(\mathbf {P})$.

$$\begin{aligned}&\begin{array}{l} \frac{5}{2} > 5 - F_D - F_{{ PD}} - F_I - F_{{ PT}} \ge \frac{3}{2} \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_{{ PD}}) \le D(\mathbf {P}, \mathbf {R}_D)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{{ PD}}) < D(\mathbf {P}, \mathbf {R}_I)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{{ PD}}) < D(\mathbf {P}, \mathbf {R}_{{ PT}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_{{ PD}}) < D(\mathbf {P}, \mathbf {R}_T) \end{array}\end{aligned}$$

(41)

$$\begin{aligned}&\begin{array}{l} \frac{7}{2} \ge F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{5}{2} \end{array} \iff \begin{array}{l} \frac{7}{2} \ge F_D + F_{{ PD}} + F_I + F_{{ PT}}\,\wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{5}{2}\,\wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > 2 \wedge \,\\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{3}{2} \end{array}\end{aligned}$$

(42)

$$\begin{aligned}&\begin{array}{l} \frac{7}{2} \ge F_D \!+\! F_{{ PD}} \!+\! F_I \!+\! F_{{ PT}} > \frac{5}{2} \end{array} \iff \begin{array}{l} \frac{7}{2} \ge F_D \!+\! F_{{ PD}} \!+\! F_I \!+\! F_{{ PT}} > \frac{5}{2} \end{array} \end{aligned}$$

(43)

Finally, we address the case of $D = {{\mathrm{MOR}}}(\mathbf {P})$.

$$\begin{aligned}&\begin{array}{l} \frac{3}{2} > 5 - F_D - F_{{ PD}} - F_I - F_{{ PT}} \ge 1 \end{array} \iff \begin{array}{l} D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_{{ PD}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_I)\,\wedge \\ D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_{{ PT}})\,\wedge \\ D(\mathbf {P}, \mathbf {R}_D) < D(\mathbf {P}, \mathbf {R}_T) \end{array}\end{aligned}$$

(44)

$$\begin{aligned}&\begin{array}{l} 4 \ge F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{7}{2} \end{array} \iff \begin{array}{l} F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{7}{2}\,\wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > 3 \,\wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{5}{2} \wedge \\ F_D + F_{{ PD}} + F_I + F_{{ PT}} > 2 \end{array} \end{aligned}$$

(45)

Because the first expression in Eq. (45) is the strictest, we can omit the remaining three. Additionally, Lemma 2 states that any component of a cumulative distribution vector is at most 1. Because of this, we can say that the sum of any four components of cumulative distribution vector is at most 4. Therefore we can safely add the upper-bound.

$$\begin{aligned} \begin{array}{l} 4 \ge F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{7}{2} \end{array} \iff \begin{array}{l} 4 \ge F_D + F_{{ PD}} + F_I + F_{{ PT}} > \frac{7}{2} \end{array} \end{aligned}$$

(46)

We have shown that equivalences in Eqs. (32), (35), (38), (41), (44) hold. This thus proves Theorem 2. $\square$