Let $S_6 = \{1, 2, 3, 4, 5, 6\}$ and define a relation $R \subseteq S_6 \times S_6$ by
$R = \{(n, m) \text{ | } n < m\text{ or }n = m + 1\}.$
Question: Write down all of the elements of $S_6 \times S_6$ which are related to $u = (6, 5)$ under $R′$, that is all the elements $v$ such that $(v, u) \in R'$.
I have written down the elements of $R$: $R=\{(1,2),(1,3),(1,4),(1,5),(1,6),(2,1),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,5),(3,6),(4,3),(4,5),(4,6),(5,4),(5,6),(6,5)\}$
It is my understanding that $R'$ means the lexicographic extension of R.
So, $(a,b)R'(c,d)$ if $aRc$ or $a=c$ and $bRd$.
Would someone oversee my logic please.
$(1,1) \in S_6 \times S_6$. Since $1R6$, $(1,1) \in R'$
$(1,2) \in S_6 \times S_6$. Since $2R6$, $(1,2) \in R'$
...
$(6,1) \in S_6 \times S_6$. Since $6=6$ and $1R5$, $(6,1) \in R'$
...
$(6,5) \in S_6 \times S_6$. Since $6=6$ and $5=5$, $(6,5) \notin R'$
$(6,6) \in S_6 \times S_6$. Since $6=6$ and $6R5$, $(6,5) \in R'$
So it is my understanding that the only ordered pair in $S_6 \times S_6 \notin R'$ is $(6,5)$. Am I correct?