$R$ is a partial order on $A$.
$B_1 \subseteq A$
$B_2 \subseteq A$
$\forall x \in B_1 \exists y \in B_2 (xRy)$
$\forall x \in B_2 \exists y \in B_1 (xRy)$
Prove that for all $x \in A$, x is an upper bound of $B_1$ iff $x$ is an upper bound of $B_2$.
My attempt:
$(\rightarrow)$ Suppose $x$ is an upper bound of $B_1$.
It means $\forall b \in B_1(bRx)$
Take arbitrary $y \in B_2$. There exists $b \in B_1$ such that $yRb$. We also know that $bRx$. Since $R$ is partial order, we have $yRx$. $y$ was arbitrary, so we have
$$\forall y \in B_2(yRx)$$
Hence $x$ is upper bound of $B_2$.
$(\leftarrow)$ (Nearly identical)
Suppose $x$ is upper bound of $B_2$. Then for all $b \in B_2$ we have $bRx$.
Take arbitrary $y \in B_1$. There exists $b \in B_2$ such that $yRb$. We also know that $bRx$. $R$ is partial order, hence $yRx$. $y$ was arbitrary, hence we conclude that $x$ is upper bound for $B_1$
$\Box$
Is it correct?