Additional Proofs

A REPULSIVE BOUNDED-CONFIDENCE MODEL

C.1 Additional Proofs

Proposition 1. Let 𝐺 = (𝑉 , 𝐸) be a network with 2 nodes , so that 𝑉 = {0,1}.

Then the dynamical process described by Equation 3.4 converges, and at time of convergence𝑇,

|𝑥₀(𝑇) −𝑥₁(𝑇) | ≤max{𝑐,|𝑥₀(0) −𝑥₁(0) |}.

Proof. Suppose that there are no edges, or 𝐸 = ∅. Then the model converges at time𝑇 =1, and

|𝑥₀(𝑇) −𝑥₁(𝑇) | =|𝑥₀(0) −𝑥₁(0) |.

Now suppose that 𝐸 = {(𝑥₀, 𝑥₁)}, and |𝑥₀(0) −𝑥₁(0) | ≥ 𝑐. Then the update rule will result in no changes, the model converges at𝑇 =1

|𝑥₀(𝑇) −𝑥₁(𝑇) | =|𝑥₀(0) −𝑥₁(0) |.

Now suppose that|𝑥₀(0) −𝑥₁(0) | < 𝑐. If𝐴₀₁=−1, the two nodes repel each other.

Then

𝑥₁(1) =𝑥₁(0) + (𝑥₁(0) −𝑥₁(0)) +𝑐− (𝑥₁(0) −𝑥₀(1)) 2

𝑥₀(1) =𝑥₀(0) + (𝑥₀(0) −𝑥₀(0)) − (𝑐− (𝑥₁(0) −𝑥₀(0)) 2

𝑥₁(1) −𝑥₀(1) =𝑥₁(0) −𝑥₀(0) + 2(𝑐− (𝑥₁(0) −𝑥₀(0))) 2

=𝑐

and𝑥₁(1) −𝑥₀(1) >=𝑐, so that after this time, these two nodes will no longer affect each other, and cannot push each other further, so the model has converged, and max𝑖, 𝑗|𝑥₀(𝑇) −𝑥₁(𝑇) | ≤𝑐.

If 𝐴₀₁ =1, the two nodes attract each other, and the model is equivalent to standard Hegselmann–Krause, so that we have convergence to a single point and

|𝑥₀(𝑇) −𝑥₁(𝑇) | =0.

This covers all possible cases, and the proposition is proven. □

Lemma 1. Suppose𝑖 ∈𝑉 a node. Define the following sets:

𝑉⁺

𝑖 (𝑡)=

𝑗 ∈𝑉 : 𝐴_{𝑖 𝑗} =1and|𝑥_𝑗(𝑡) −𝑥_𝑖(𝑡) | < 𝑐 𝑈_𝑖(𝑡)=

𝑗 ∈𝑉 : 𝐴_{𝑖 𝑗} =−1and

(0< 𝑥_𝑗(𝑡) −𝑥_𝑖(𝑡) < 𝑐)or 𝑥_𝑗(𝑡) =𝑥_𝑖(𝑡) and 𝑗 > 𝑖 𝐿_𝑖(𝑡)=

𝑗 ∈𝑉 : 𝐴_{𝑖 𝑗} =−1and

(0< 𝑥_𝑖(𝑡) −𝑥_𝑗(𝑡) < 𝑐)or 𝑥_𝑗(𝑡) =𝑥_𝑖(𝑡) and𝑖 > 𝑗 .

Then

𝑥_𝑖(𝑡+1) = Í

𝑗∈𝑉⁺

𝑖 (𝑡)𝑥_𝑗(𝑡) +Í

𝑗∈𝑈𝑖(𝑡)(𝑥_𝑗(𝑡) −𝑐) +Í

𝑗∈𝐿𝑖(𝑥_𝑗(𝑡) +𝑐)

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) + |𝐿_𝑖(𝑡) | . (3.5) Proof. From Equation 3.4, we rearrange

𝑥_𝑖(𝑡+1) =𝑥_𝑖(𝑡) + Í

𝑗∈𝑉 𝐴_{𝑖 𝑗}𝑀_{𝑖 𝑗}(𝑡)1_|𝑥 𝑗(𝑡) −𝑥

𝑖(𝑡) |<𝑐

Í

𝑗∈𝑉|𝐴_{𝑖 𝑗}|1_|𝑥_𝑗(𝑡) −𝑥_𝑖(𝑡) |<𝑐

=𝑥_𝑖(𝑡) + Í

𝑗∈𝑉⁺

𝑖 (𝑡)∪𝑈_𝑖(𝑡)∪𝐿_𝑖(𝑡) 𝐴_{𝑖 𝑗}𝑀_{𝑖 𝑗}(𝑡) Í

𝑗∈𝑉⁺

𝑖(𝑡)∪𝑈_𝑖(𝑡)∪𝐿_𝑖(𝑡)|𝐴_{𝑖 𝑗}|

=𝑥_𝑖(𝑡) + Í

𝑗∈𝑉⁺

𝑖(𝑡)(𝑥_𝑗(𝑡) −𝑥_𝑖(𝑡))

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) | + |𝐿_𝑖(𝑡) | +

Í

𝑗∈𝑈_𝑖(𝑡)(−1) (1) (𝑐− (𝑥_𝑗(𝑡) −𝑥_𝑖(𝑡)))

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) | + |𝐿_𝑖(𝑡) | +

Í

𝑗∈𝐿_𝑖(𝑡)(−1) (−1) (𝑐− (𝑥_𝑖(𝑡) −𝑥_𝑗(𝑡)))

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) | + |𝐿_𝑖(𝑡) |

= Í

𝑗∈𝑉⁺

𝑖 (𝑡)𝑥_𝑗(𝑡) +Í

𝑗∈𝑈_𝑖(𝑡)(𝑥_𝑗(𝑡) −𝑐) +Í

𝑗∈𝐿_𝑖(𝑥_𝑗(𝑡) +𝑐)

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) + |𝐿_𝑖(𝑡) | .

□ Lemma 2. Let𝑖 ∈𝑉 at time𝑡, and let𝑊(𝑡) ⊂𝑉 be a set of nodes such that𝑊(𝑡)is completely contained in𝑉⁺

𝑖 (𝑡) ∪𝑈_𝑖(𝑡) ∪𝐿_𝑖(𝑡). Define 𝑊(𝑡)=

Í

𝑗∈𝑊(𝑡)𝑥_𝑗(𝑡)

|𝑊(𝑡) |

to be the average of𝑥_𝑗(𝑡) for all 𝑗 ∈𝑊(𝑡). Then we can rewrite Equation3.5as

𝑥_𝑖(𝑡+1) = Í

𝑗∈(^𝑉𝑖⁺(𝑡)∪𝑈_𝑖(𝑡)∪𝐿_𝑖(𝑡))^\𝑊(𝑡)𝑥_𝑗(𝑡) +Í

𝑗∈𝑊(𝑡)𝑊(𝑡) + (|𝐿_𝑖(𝑡) | − |𝑈_𝑖(𝑡) |)𝑐

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) + |𝐿_𝑖(𝑡) | .

Proof. We rearrange Equation 3.5 as follows:

𝑥_𝑖(𝑡+1) = Í

𝑗∈𝑉⁺

𝑖 (𝑡)𝑥_𝑗(𝑡) +Í

𝑗∈𝑈𝑖(𝑡)(𝑥_𝑗(𝑡) −𝑐) +Í

𝑗∈𝐿𝑖(𝑥_𝑗(𝑡) +𝑐)

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) + |𝐿_𝑖(𝑡) |

= Í

𝑗∈(^𝑉𝑖⁺(𝑡)∪𝑈_𝑖(𝑡)∪𝐿_𝑖(𝑡))𝑥_𝑗(𝑡) + (|𝐿_𝑖(𝑡) | − |𝑈_𝑖(𝑡) |)𝑐

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) + |𝐿_𝑖(𝑡) |

= Í

𝑗∈(^𝑉𝑖⁺(𝑡)∪𝑈_𝑖(𝑡)∪𝐿_𝑖(𝑡))^\𝑊(𝑡)𝑥_𝑗(𝑡) +Í

𝑗∈𝑊(𝑡)𝑊(𝑡) + (|𝐿_𝑖(𝑡) | − |𝑈_𝑖(𝑡) |)𝑐

|𝑉⁺

𝑖 (𝑡) | + |𝑈_𝑖(𝑡) + |𝐿_𝑖(𝑡) | .

□ Lemma 3. Let𝐺 = (𝑉 , 𝐸)be a network with𝑛nodes and𝑚edges with confidence bound𝑐. Suppose that every edge in𝐺 is repulsive. At time𝑡, suppose𝑥_𝑖(𝑡) > 𝑥_𝑗(𝑡) for all other nodes 𝑗, so that𝑖 is the node with the highest opinion value at time𝑡. Then𝑥_𝑖(𝑡+1) > 𝑥_𝑗(𝑡+1)for all 𝑗.

Proof. Note that𝑉⁺

𝑘(𝑡) = {𝑥_𝑘} for all 𝑘 , 𝑡, since every edge in𝐺 is repulsive. For convenience we define the following sets:

𝑈_{𝑖 𝑗}(𝑡)=𝑈_𝑗(𝑡)Ù 𝑈_𝑖(𝑡) 𝑈^′

𝑖 𝑗(𝑡)=𝑈_𝑖(𝑡) \𝑈_{𝑖 𝑗}(𝑡) 𝐿_{𝑖 𝑗}(𝑡)= 𝐿_𝑖(𝑡)Ù

𝐿_𝑗(𝑡) 𝐿^′

𝑗 𝑖(𝑡)= 𝐿_𝑗(𝑡) \𝐿_{𝑖 𝑗}(𝑡) 𝑊_{𝑖 𝑗}(𝑡)= 𝐿_𝑖(𝑡)Ù

𝑈_𝑗(𝑡).

Unpacking this notation,𝑈_{𝑖 𝑗}(𝑡)consists of all nodes that repel both𝑖and𝑗downward, while 𝐿_{𝑖 𝑗}(𝑡) consists of all nodes that repel both𝑖 and 𝑗 upward. 𝑈^′

𝑖 𝑗(𝑡) consists of nodes which repel 𝑖 downward, but not 𝑗 (note that if 𝑥_𝑖(𝑡) < 𝑥_𝑗(𝑡), this is automatically empty), while𝐿^′

𝑗 𝑖(𝑡)consists of nodes which repel 𝑗 upward, but not 𝑖 (again, if 𝑥_𝑖(𝑡) < 𝑥_𝑗(𝑡), this is empty). Finally, 𝑊_{𝑖 𝑗}(𝑡) consists of nodes which repel𝑖upward and 𝑗 downward (empty if𝑥_𝑗(𝑡) > 𝑥_𝑖(𝑡)).

Now, suppose𝑥_𝑖(𝑡) > 𝑥_𝑗(𝑡)for all 𝑗 ∈𝑉. Then we can write 𝑉⁺

𝑖 (𝑡) ={𝑖} 𝑈_𝑖(𝑡) =∅

𝐿_𝑖(𝑡) =𝐿_{𝑖 𝑗}(𝑡)Ø

𝑊_{𝑖 𝑗}(𝑡)Ø {𝑗}

and

𝑉⁺

𝑗(𝑡) ={𝑗} 𝑈_𝑖(𝑡) ={𝑖}Ø

𝑊_{𝑖 𝑗}(𝑡) 𝐿_𝑖(𝑡) =𝐿_{𝑖 𝑗}(𝑡)Ø

𝐿^′

𝑗 𝑖(𝑡).

Then we observe the following from the knowledge that nodes only effect each other if they are within confidence of each other.

𝑥_𝑗(𝑡) +𝑐 > 𝑥_𝑖(𝑡) 𝐿_{𝑖 𝑗}(𝑡) +𝑐 > 𝑥_𝑖(𝑡) 𝑊_{𝑖 𝑗}(𝑡) +𝑐 > 𝑥_𝑖(𝑡)

𝑥_𝑖(𝑡) > 𝐿^′

𝑗 𝑖+𝑐

Then applying Lemma 1 and Lemma 2:

𝑥_𝑖(𝑡+1)

=

𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝑊_{𝑖 𝑗}(𝑡) | (𝑊_{𝑖 𝑗}(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝑊_{𝑖 𝑗}(𝑡) |

>

𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝑊_{𝑖 𝑗}(𝑡) | (𝑊_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝑊_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

(C.1)

>

(𝑥_𝑖(𝑡) −𝑐) +𝑥_𝑗(𝑡) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝑊_{𝑖 𝑗}(𝑡) | (𝑊_{𝑖 𝑗}(𝑡) −𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝑊_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

=𝑥_𝑗(𝑡+1)

where the inequality in Equation C.1 follows because𝑥_𝑖(𝑡+1)is a weighted average, and 𝐿^′

𝑗 𝑖(𝑡) is less than all of the other values being averaged in the previous line.

The next inequality follows straightforwardly by replacing each value in the average with a smaller or equal value.

So if𝑥_𝑖(𝑡) has the highest value opinion at time𝑡, it will always have the highest

value opinion. □

Corollary 1. Let𝐺 =(𝑉 , 𝐸)be a network with𝑛nodes and𝑚edges with confidence bound𝑐. Suppose that every edge in𝐺 is repulsive. At time𝑡, let 𝑀 = {𝑖 : 𝑥_𝑖(𝑡) ≥ 𝑥_𝑗(𝑡)∀𝑗 ∈𝑉}. Then𝑥_max

𝑀𝑖(𝑡+1) > 𝑥_𝑗(𝑡+1)∀𝑗 ∈𝑉.

Proof. From the definitions of𝑈_𝑖(𝑡), 𝐿_𝑖(𝑡), we can observe that the member of 𝑀 with highest index will have the largest corresponding set 𝐿_𝑖(𝑡) and the smallest corresponding𝑈_𝑖(𝑡), so that at time𝑡+1, that member of 𝑀 will have the highest- valued opinion of all members of𝑀. By the same logic as in the proof of Lemma 3, that opinion will also be the highest-valued opinion overall. □ Corollary 2. Let𝐺 = (𝑉 , 𝐸)be the complete network with𝑛nodes with confidence bound𝑐. Suppose that every edge in𝐺 is repulsive. At time𝑡, let 𝑀 = {𝑖 : 𝑥_𝑖(𝑡) ≤ 𝑥_𝑗(𝑡)∀𝑗 ∈𝑉}. Then𝑥_min

𝑀𝑖(𝑡+1) < 𝑥_𝑗(𝑡+1)∀𝑗 ∈𝑉.

Proof. Proves that𝑥_𝑖(𝑡) < 𝑥_𝑗(𝑡)for all 𝑗 ∈𝑉, then𝑥_𝑖(𝑡+1) < 𝑥_𝑗(𝑡+1)for all 𝑗 ∈𝑉, by segmenting𝑈_𝑖(𝑡), 𝐿_𝑖(𝑡), 𝑈_𝑗(𝑡), 𝐿_𝑗(𝑡) into the appropriate subsets and reversing inequalities as needed as in lemma 3. Then the same logic as in corollary 1 proves

the statement. □

Lemma 4. Let𝐺 = (𝑉 , 𝐸)be a network with𝑛nodes and𝑚edges with confidence bound𝑐. Suppose that every edge in𝐺 is repulsive. At time𝑡, suppose𝑥_𝑖(𝑡) > 𝑥_𝑗(𝑡) for all other nodes 𝑗 ∈ 𝑉, so that𝑖 is the node with the highest-valued opinion at time𝑡. Suppose that there is some node 𝑗 such that𝑥_𝑖(𝑡) −𝑥_𝑗(𝑡) < 𝑐, and that 𝑗 has the highest-valued opinion of all such nodes. Then

2𝑐 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | ≤ 𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1) ≤

(|𝐿^′

𝑗 𝑖(𝑡) | +2)𝑐 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |. Proof. By assumption, since 𝑗 has the highest-valued opinion of all nodes within confidence of𝑖,𝑊_{𝑖 𝑗}(𝑡)is empty. To prove the lower bound,

𝑥_𝑖(𝑡+1) =

𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) |

≥

𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | 𝑥_𝑗(𝑡+1) =

(𝑥_𝑖(𝑡) −𝑐) +𝑥_𝑗(𝑡) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | 𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1) ≥ 2𝑐

2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |.

To prove the upper bound, 𝑥_𝑖(𝑡+1) =

𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐)+

2+ |𝐿_{𝑖 𝑗}(𝑡) |

≤

𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝑥_𝑗(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | 𝑥_𝑗(𝑡+1) =

(𝑥_𝑖(𝑡) −𝑐) +𝑥_𝑗(𝑡) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | 𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1) ≤

𝑐+𝑐+ |𝐿^′

𝑗 𝑖(𝑡) | (𝑥_𝑗(𝑡) −𝐿^′

𝑗 𝑖(𝑡)) 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

≤

(|𝐿^′

𝑗 𝑖(𝑡) | +2)𝑐 2+ |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |. Notice that if 𝐿^′

𝑗 𝑖(𝑡) is empty, both inequalities become equalities, so that 𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1)= 2𝑐

2+ |𝐿_{𝑖 𝑗}(𝑡) | Notice also that if both 𝐿_{𝑖 𝑗}(𝑡) and 𝐿^′

𝑗 𝑖(𝑡) are empty, that the distance between

𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1) is precisely𝑐. □

Corollary 3. Let𝐺 =(𝑉 , 𝐸)be a network with𝑛nodes and𝑚edges with confidence bound𝑐. Suppose that every edge in𝐺 is repulsive. At time𝑡, suppose𝑥_𝑖(𝑡) < 𝑥_𝑗(𝑡) for all other nodes 𝑗 ∈𝑉, so that𝑖is the node with the lowest-valued opinion at time 𝑡. Suppose that there is some node 𝑗 such that𝑥_𝑗(𝑡) −𝑥_𝑖(𝑡) < 𝑐, and that 𝑗 has the lowest-valued opinion of all such nodes. Then

2𝑐 2+ |𝑈_{𝑖 𝑗}(𝑡) | + |𝑈^′

𝑖 𝑗(𝑡) | ≤ 𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1) ≤

(|𝑈^′

𝑖 𝑗(𝑡) | +2)𝑐 2+ |𝑈_{𝑖 𝑗}(𝑡) | + |𝑈^′

𝑖 𝑗(𝑡) |. Lemma 5. Let𝐺 = (𝑉 , 𝐸) be the complete network with 𝑛nodes and confidence bound𝑐. Suppose that every edge in𝐺 is repulsive. At time𝑡, suppose that𝑖and 𝑗 are nodes such that(𝑖, 𝑗) ∈ 𝐸,𝑥_𝑖(𝑡) > 𝑥_𝑗(𝑡), and𝑥_𝑖(𝑡) −𝑥_𝑗(𝑡) < 𝑐, and there exist no nodes𝑘 connected to𝑖or 𝑗 such that𝑥_𝑖(𝑡) > 𝑥_𝑘(𝑡) > 𝑥_𝑗(𝑡). Then

|𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1) | ≤𝑐 .

Proof. By the assumption that no nodes have values between𝑥_𝑖(𝑡) and 𝑥_𝑗(𝑡), we have that𝑊_{𝑖 𝑗}(𝑡)=𝑊_{𝑗 𝑖}(𝑡) =∅. Then to prove one direction of the bound,

𝑥_𝑖(𝑡+1)=

|𝑈^′

𝑖 𝑗(𝑡) | (𝑈^′

𝑖 𝑗(𝑡) −𝑐) + |𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) +𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) |

≤

|𝑈^′

𝑖 𝑗(𝑡) | (𝑈^′

𝑖 𝑗(𝑡) −𝑐) + |𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) +𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | +

|𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝑥_𝑗(𝑡) +𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | 𝑥_𝑗(𝑡+1)=

|𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) + (𝑥_𝑖(𝑡) −𝑐) +𝑥_𝑗(𝑡) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

≥

|𝑈^′

𝑖 𝑗(𝑡) | (𝑥_𝑖(𝑡) −𝑐) + |𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) + (𝑥_𝑖(𝑡) −𝑐) +𝑥_𝑗(𝑡) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | +

|𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |.

Combining both equations, 𝑥_𝑖(𝑡+1) −𝑥_𝑗(𝑡+1) ≤

|𝑈^′

𝑖 𝑗(𝑡) | 𝑈^′

𝑖 𝑗(𝑡) −𝑥_𝑖(𝑡)

+𝑐+𝑐+ |𝐿^′

𝑗 𝑖(𝑡) |

𝑥_𝑗(𝑡) −𝐿^′

𝑗 𝑖(𝑡) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

≤

2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | 𝑐 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

≤ 𝑐 . To prove the other direction,

𝑥_𝑗(𝑡+1)=

|𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) + (𝑥_𝑖(𝑡) −𝑐) +𝑥_𝑗(𝑡) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

≤

|𝑈^′

𝑖 𝑗(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) + (𝑥_𝑖(𝑡) −𝑐) +𝑥_𝑗(𝑡) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | +

|𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝐿^′

𝑗 𝑖(𝑡) +𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | 𝑥_𝑖(𝑡+1)=

|𝑈^′

𝑖 𝑗(𝑡) | (𝑈^′

𝑖 𝑗(𝑡) −𝑐) + |𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) +𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) + |𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) |

≥

|𝑈^′

𝑖 𝑗(𝑡) | (𝑈^′

𝑖 𝑗(𝑡) −𝑐) + |𝑈_{𝑖 𝑗}(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) +𝑥_𝑖(𝑡) + (𝑥_𝑗(𝑡) +𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | +

|𝐿_{𝑖 𝑗}(𝑡) | (𝐿_{𝑖 𝑗}(𝑡) +𝑐) + |𝐿^′

𝑗 𝑖(𝑡) | (𝑈_{𝑖 𝑗}(𝑡) −𝑐) 2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |.

Combining both inequalities yields

𝑥_𝑗(𝑡+1) −𝑥_𝑖(𝑡+1) ≤

|𝑈^′

𝑖 𝑗(𝑡) |

𝐿_{𝑖 𝑗}(𝑡) −𝑈^′

𝑖 𝑗(𝑡) +2𝑐

+𝑐+𝑐+ |𝐿^′

𝑗 𝑖(𝑡) | 𝐿^′

𝑗 𝑖(𝑡) −𝑈_{𝑖 𝑗}(𝑡) +2𝑐

2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) | .

Note, however, that

2𝑐= (𝑥_𝑖(𝑡) +𝑐) − (𝑥_𝑖(𝑡) −𝑐)

> 𝑈^′

𝑖 𝑗(𝑡) −𝐿_{𝑖 𝑗}(𝑡)

> (𝑥_𝑗(𝑡) +𝑐) −𝑥_𝑗(𝑡) =𝑐 and similarly𝑐 < 𝑈_{𝑖 𝑗}(𝑡) −𝐿^′

𝑗 𝑖(𝑡) < 2𝑐so that we have 𝑥_𝑗(𝑡+1) −𝑥_𝑖(𝑡+1) ≤

|𝑈^′

𝑖 𝑗(𝑡) |

𝐿_{𝑖 𝑗}(𝑡) −𝑈^′

𝑖 𝑗(𝑡) +2𝑐

+𝑐+𝑐+ |𝐿^′

𝑗 𝑖(𝑡) |

𝐿^′

𝑗 𝑖(𝑡) −𝑈_{𝑖 𝑗}(𝑡) +2𝑐

2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

≤

2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

𝑐

2+ |𝑈^′

𝑖 𝑗(𝑡) | + |𝑈_{𝑖 𝑗}(𝑡) | + |𝐿_{𝑖 𝑗}(𝑡) | + |𝐿^′

𝑗 𝑖(𝑡) |

≤ 𝑐

and the proof is finished. □

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