Question: Show that if $R$ is a strict partial order on $X$, and $R$ is not linear, then there exists a strict partial order $R'$ and $R' \supsetneqq R$.
My attempt: By definition 6.23,6.3.1, and 6.3.2ii,
R is irreflexive if $(\forall x \in S[(x,x) \notin R)$
R is antisymmetric if $(\forall x,y \in S)((x,y) \in R \land (y,x) \in R] \rightarrow (x=y)]$
R is transitive if $( \forall x,y,z \in S)[(x,y) \in R \land (y,z) \in R \rightarrow (x,z) \in R)]$
Furthermore, by Definition 6.3.2iii, a linear order occurs if $R$ is a strict partial order and every two elements of $S$ are comparable.
Suppose $R$ isn't a linear order. Then $R$ isn't a strict partial order, and there is at least one pair of elements that aren't comparable. Also, $R'$ means that it's reflexive, symmetric, and not transitive.
Here's my question...since I've already given the definition of linear and strict partial order, what would be a good example to show that $R' \supsetneqq R$. Maybe there's a unique element that exists in $R'$, but not in $R$. What if I let $R =$ all 26 letters of the alphabet and $R' =$ all 26 letters of the alphabet and numbers?
I tried to let $R=1,2,3$ and $R'=3,4,5$ but that didn't work since all of the same elements must be in $R$ and $R'$ meaning there must be $R =1,2,3$ and $R' = 1,2,3$ . However there is nothing unique. Sure all of the elements are in $R$ and $R'$, but where is the uniqueness in $R'$? So I thought about the alphabets and numbers which kind of work but seem childish.
What if I let the example be that $R = 1,2,3$ and $R' = 1,2,3,4$? There are elements in common in $R$ and $R'$ which is $1,2,3$, but $R'$ only has the number 4. As a result $R' \supsetneqq R$. What do you guys think? Any improvements? hints?
Let $R$ be a strict partial order on $S. \ $ For ex. $R = \{(x,y)\}, \ $ where $x \neq y$ are from $S$, is a strict partial order. Let $R' = \{(y,x)\}$. Clearly that is another SPO, and disjoint from $R$. How did we come about it? Well, we reversed the relation i.e. if $(x,y) \in R$, then put $(y, x) \in R'$. This might work in the general case. Let's check.
Let $R$ be an SPO. Let $R' = R \text{ with relationships reversed}. \ $ Then Clearly $R'$ is irreflexive, and transitive (check this one). And the definition to antisymmetry is the same wth $x,y$ reversed (as text in the definition!!!), so is also satisfied. And $R \cap R' = \varnothing$ or else $(x,y) \in R \cap R' \implies (y,x) \in R'$ and by transitivity $(x,x) \in R'$ which violates irreflexivity. Thus we've done way better than $R' \supsetneq R$.
That proves what you set out to.