Relation $R_1$ on set $A$ and Relation $R_2$ on set $B$ are defined.
$R_1$ and $R_2$ are both partial orders.
We define $R$ like below:
$$(a, b) R (x, y) \quad\text{ if }\quad aR1x \;\text{ and }\; bR2y$$
Prove that $R$ is also a partial order.
I know that a relation $R$ on a set $S$ is called a partial order if it is reflexive, anti-symmetric, and transitive.
What the problem requires of you is to manually check if the three partial order properties are satisfied.
Reflexivity: $(a,b)R(a,b)$ is in $R$ since $aR_1a$ and $bR_2b$ are satisfied due to their being partial orders.
Anti-symmetry: suppose we have $(a,b)R(c,d)$ and $(c,d)R(a,b)$. Then $aR_1c$ and $cR_1a$, which implies $a=c$ by anti-symmetry of $R_1$, and similarly for $R_2$. Hence $(a,b)=(c,d)$, as required.
Transitivity: let $(a,b)R(c,d)$ and $(c,d)R(e,f)$. Then $aR_1c$ and $cR_1e$, which implies $aR_1e$ by transitivity of $R_1$, and similarly for $R_2$. Hence we have $(a,b)R(e,f)$, as required.