Proof verification: each $R_i$ reflexive\complete and transitive then so is $R$ defined as $xRy$ if and only if $xR_iy,\forall i \in \{1,\ldots,n\}$. Suppose $R_1, R_2,\ldots,R_n$ are binary relations on $X$. Define the binary relation $R$ by
$$xRy\quad\text{if and only if}\quad xR_iy,\forall i \in \{1,\ldots,n\}.$$
Prove or provide counterexamples to the following statements:
(a) If each $R_i$ is reflexive and transitive, then $R$ is reflexive and transitive.
(b) If each $R_i$ is complete and transitive, then $R$ is complete and transitive.
I couldn't think of a counter example to these two statements so I have tried to use the definitions to prove, but I am not yet very confident in my abilities so I am asking for proof verification.
My attempt:
(a) Suppose $R$ is not reflexive or transitive, then $x\not Rx$ for some $x$ s.th. $x \not R_i x$ for some $i\in\{1,\ldots,n\}$. But this is not possible as $xR_iy \forall i\in\{1,\ldots,n\}$. Hence R must be reflexive and transitive.
(b) Let $X=\{x,y,z\}$, then if $xR_iy, yR_iz, zR_ix$ $\forall R_i$ then by definition of $R$, $xRy, yRz$ and $xRz$, s.th. $R$ is complete and transitive.

(a) This is true. You want to prove that $R$ is reflexive and transitive. In fact:

*

*If $x\in X$, then, for each $i\in\{1,2,\ldots,n\}$, $x\mathrel{R_i}x$, and therefore $x\mathrel Rx$. So, $R$ is reflexive.

*If $x,y,z\in X$, and $x\mathrel Ry$ and $y\mathrel Rz$, you want to prove that $x\mathrel Rz$. You are assuming that, for each $i\in\{1,2,\ldots,n\}$, $x\mathrel{R_i}y$ and $y\mathrel{R_i}z$. Therefore, again for each $i\in\{1,2,\ldots,n\}$, $x\mathrel{R_i}z$. It follows that $x\mathrel Rz$. So, $R$ is indeed transitive.

(b) This is false. Take $X=\{0,1\}$, $R_1=\leqslant$, and $R_2=\geqslant$. Then both $R_1$ and $R_2$ are complete and transitive. However, the binary relation $R$ is simply the equality on $X$, which is not complete.
