Proving the uniqueness of a function $r-s$-order preserver in $A$ and $B$.

Question

Definitions Let A, B sets

1. $r$- well ordered to $A$ if $\forall u,v \in A$ one and just one of the following afirmation is true $urv, vru, u=v.$ and $\forall B \subset A, \exists v \in B$ such that $\forall u \in B, urv$ is false.

2. $B$ is a $r$-section of A if $B \subset A$, $r$-well ordered to $A$ and $\forall u \in A, v \in B$ then $u \in B$.

3. $f$ is a function if $(x,y) \in f$ and $(x,z) \in f$ then $y=z$ equivalently $f= \{(x,y) : f(x)=y\}$.

4. $f$ is $r-s$-order preserver in $A$ and $B$ if $r$-well ordered to $domf$, $s$-well ordered to $ran f$, $\forall u,v \in domf$, $f(urv)$ then $f(u)sf(v)$, $domf$ is a $r$-section of $A$ and $ranf$ is a $s$-section of $B$.

Problem If $r$-well ordered to $A$ and $s$-well ordered $B$ then there exists at most one function $f$ $r-s$-order preserver in $A$ and $B$ such that $domf=A$ or $ranf=B$.

What I have is the existence of that function but I have problems to prove uniqueness. Could you help me? please.

Answer 1

HINT: Suppose that $f$ and $g$ are two such functions, and let $$C=\{a\in A:f(a)\ne g(a)\}\,.$$ If $C\ne\varnothing$, there is a $c_0\in C$ such that for each $c\in C$, $crc_0$ is false. (In more intuitive terms, $c_0$ is the $r$-smallest element of $C$.) Without loss of generality suppose that $f(c_0)sg(c_0)$. Get a contradiction by showing that $\operatorname{ran}g$ is not an $s$-section of $B$.