Define the relation $\le$, as $(a,b)\le(c,d)$ if and only if $a+b\le c+d$ and $a\le c$.
Is this a partial order?
I know it's definitely not if we remove the $a\le c$ (because then it's not antisymmetric - example (1,3) and (2,2)), but not sure how to show one way or another with that $a\le c$ tacked on. At first I thought it might simply be a weird way to write the lexicographic order, but that doesn't seem to be true. Any help is appreciated.
On
Reflexivity: Let $(a,b) \in R$. Then $(a,b) \leq (a,b)$ since a+b= a+b $\rightarrow$ a+b $\leq$ a+b and clearly $a\leq a$.
Antisymmetry: Let $(a,b) \leq (c,d)$ and $(c,d) \leq (a,b)$. Then $a+b \leq c+d $ and $a\leq c$. Also, $c+d \leq a+b$ and $c\leq a$. Clearly, c < a and a < c is a contradiction unless a= c. By the same reasoning since a=c,a+b < c+d and c+d < a+b is a contradiction unless a+b = c+d. But since a=c, this implies b=d.So by the definition of equal ordered pairs, $(a,b)= (c,d)$.
Transitivity: This is a little tedious. Let $(a,b) \leq (c,d)$ and $(c,d) \leq (e,f)$. Then $a+b \leq c+d $ and $a\leq c$. Also, $c+d \leq e+f $ and $c\leq e$. Then $a\leq c \leq e$ $\rightarrow$ $a\leq e$. So $a+b \leq c+d \leq e+f $ $\rightarrow$ $a+b \leq e+f $. So $(a,b) \leq (e,f)$ and this gives transitivity and we're done!
To show that $\leq$ is a partial order, we need to show three things:
The first definitely holds, as $a \leq a$ and $a + b \leq a + b$.
The second holds: Assume $(a,b) \leq (c,d)$ and $(c,d) \leq (a,b)$. Then we have $a \leq c$ and $c \leq a$. These two imply that $a = c$. Additionally, we have $a + b \leq c + d$ and $c + d \leq a + b$. These two imply that $a + b = c + d$. However, since we already know that $a = c$, we can subtract $a$ from the left and $c$ from the right to see that $b = d$. Thus, we have $(a,b) = (c,d)$.
Finally, the last holds; $(a,b) \leq (c,d) \leq (d,e)$ implies that $a \leq c \leq d$. Moreover, $a + b\leq c + d \leq e + f$. Thus $\leq$ is a partial order.