Proof

We begin with some auxiliary definitions and we split proof into few simple lemmas. We fix a voting procedure $F$.

Definition Let $a,b \in X$ and $M\subseteq\{1,\ldots,n\}$. The set $M$ is weakly decisive for $(a,b)$ if for each $R_1,\ldots,R_n \in \mathcal{L}$, if $$ (\forall i\in M)(a R_i b) \land (\forall i\in M)(b R_i a) $$ then $a F(R_1,\ldots,R_n) b$.

Observe that the set $\{1,\ldots,n\}$ is weakly decisive for any pair $(a,b)$.

Lemma 1. Suppose that $M$ is weakly decisive for $(a,b)$. Let $x\in X\setminus\{a,b\}$. Then $M$ is weakly decisive for $(a,x)$ and is weakly decisive for $(x,b)$

Proof. Suppose that all voters from $M$ have the following ordering $a \prec b \prec x$ (restricted to $\{a,b,x\}$) and all voters from $M^c$ have ordering $b \prec x \prec a$. Let $\prec^*$ be the ordering obtained from function $F$ applied to these orderings. Fromm rule R1 we get $b \prec^* x$. From asuumptio on $M$ we get $a\prec^* b$. Therefore $a \prec^* x$. Hence $M$ is weakly decisive for $(a,x)$.
The remaining argument is symmetric. We consider the orderings $x\prec a \prec b$ for members of $M$ and the orderings $b \prec x \prec a$ for members of $M^c$. From R1 we get $x\prec^* a$ and from assumptions we get $a \prec^* b$. Therefore $x\prec^* b$. Hence $M$ is weakly decisive for $(x,b)$.

Remark. The first part of this proof can be summarized as follows: $$ \left\{ \begin{array}{rl} M: & a \prec b \prec x\\ M^c: & b \prec x \prec a \end{array} \right. \Rightarrow \left([R1] (b\prec^*x) \land [M] (a\prec^* b)\right) \Rightarrow a \prec^* x $$ [R1] means that Pareto rule was used to deduce this inequality and [M] means that the assumption about M was used.
Exercise. Draw a similar sketch of the proof of the second part of this proof and do the same for next lemmas.

Lemma 2. Suppose that $M$ is weakly decisive for $(a,b)$. Let $x\in X\setminus\{a,c\}$. Then $M$ is weakly decisive for $(a,x)$, $(x,a)$, $(x,b)$, $(b,x)$ and $(b,a)$.

Proof. Let $w(u,v)$ be an abbreviation of sentence "M is weakly decisive for $(u,v)$". We will apply Lemma 1 several times.

1. $w(a,b) \rightarrow w(a,x)$
2. $w(a,b) \rightarrow w(x,b) \rightarrow w(x,a) \rightarrow w(b,a) \rightarrow w(b,x)$

Lemma 3. Suppose that $M$ is weakly decisive for $(a,b)$. Then $M$ is weakly decisive for any other pair $(x,y)$.

Proof. Let $x,y$ be any different pair of elements from the set $X$. If $\{x,y\} = \{a,b\}$ or $(x\in\{a,b\} \land y \notin\{a,b\}$) then we may apply Lemma 2 directly. Suppose hence that $\{x,y\}\cap\{a,b\} = \emptyset$. Using the abbreviation from the previous Lemma we have $$ w(a,b) \rightarrow w(a,y) \rightarrow w(x,y)~. $$

Definition Let $a,b \in X$ and $M\subseteq\{1,\ldots,n\}$. The set $M$ is decisive for $(a,b)$ if for each $R_1,\ldots,R_n \in \mathcal{L}$, if $$ (\forall i\in M)(a R_i b) $$ then $a F(R_1,\ldots,R_n) b$.

Lemma 4. Suppose that $M$ is weakly decisive for $(a,b)$. Then $M$ is decisive for $(a,b)$.

Proof. Let $x \in X \setminus \{a,b\}$. We know that $M$ is weakly decisive for $(x,b)$.
Suppose that all voters from $M$ have the following ordering $a \prec x \prec b$ and all voters from $M^c$ have ordering $u_1 \prec u_2 \prec x$, where $(u_1,u_2)=(a,b)$ or $(u_1,u_2)=(b,a)$.
Let $\prec^*$ be the ordering obtained from function $F$ applied to these orderings. Then $x \prec^* b$. From R1 we get $b\prec^* x$. Therefore $b \prec^* a$.

Corollary. The following sentences are equivalent:

1. $(\exists a,b\in X)(M$ is weakly decisive for $(a,b))$
2. $(\forall a,b\in X)(M$ is weakly decisive for $(a,b))$
3. $(\forall a,b\in X)(M$ is decisive for $(a,b))$

We say that $M$ is decisive if there exists a pair $(a,b)$ such that $M$ is decisive for $(a,b)$. Note that this is equivalent to say that $M$ is decisive for every pair $(a,b)$.

Lemma 5. Suppose that $M$ is decisive and that $M=M_1\cup M_2$, where $M_1\cap M_2=\emptyset$, $M_1\neq\emptyset$ and $M_2\neq\emptyset$. Then $M_1$ or $M_2$ is decisive.

Proof. Let us fix three pairwise different elements $x,y,z \in X$. We put the following orderings:

1. $z \prec y \prec x$ on $M_1$
2. $x \prec z \prec y$ on $M_2$
3. $y \prec x \prec z$ on $(M_1\cup M_2)^c$

1. $z \prec x$: then $M_1$ is decisive for $(z,x)$
2. $x \prec z$: then we have $x \prec^* y$, so $M_2$ is decisive for $(x,y)$

From Lemma 5 (applied several times) we deduce that there exists $i$ such that $\{i\}$ is decisive. And this means that $\{i\}$ is decisive for all pairs $(a,b)$. This means that preferences of $i$ decides about the voting procedure. Voter $i$ may be treated as dictator.