Show that $R$ is an order relation in $\mathbb{R^2}$

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In $\mathbb{R^2}$ consider the relation $R$ as: $(x,y)R(x',y') \Longleftrightarrow x\leq x' and\ y \leq y'$

show that $R$ is an order relation.

1) $ \forall (x,y) \in \mathbb{R^2} \ (x,y)R(x,y) \implies x \leq x \ and \ y \leq y$

which is true because $x = x$ and $ y = y$ $ \implies $ $R$ is reflexive.

2) $( \ (x,y)R(x',y') \implies x \leq x' \ and \ y' \leq y \ )$ and $( \ (x',y')R(x,y) \implies x \geq x' \ and \ y' \geq y \ )$

which implies $x = x'$ and $ y = y' $ $\implies$ $R$ is anti-symmetric

3) $( \ (x,y)R(x',y') \implies x \leq x' \ and \ y' \leq y \ )$ and $( \ (x',y')R(z,z') \implies x' \leq z \ and \ y' \leq z' \ )$ $\implies x \leq z \ and \ y \leq z'$

which implies that $R$ is transitive

So $R$ is an order relation.

is this answer correct ?