**Proof:** Let *A* and *B* be two posets . We have to show that *A × B* = {(a, b) | a ∈ *A*, b ∈ *B*}

Forms a poset under the relation defined by (a_{1}, b_{1}) ≤ (a_{2}, b_{2})

**⇔ a _{1} ≤ a_{2 }in A and b_{1} ≤ b_{2} in B**

It is clear that the tree relations ≤ occurring above are different, being the respective relations in *A × B, A *and *B*.

**Reflexivity:** (a, b) ≤ (a, b) ∀ (a, b) ∈ *A × B*

as a ≤ a in *A* and b ≤ b in *B* ∀ a ∈ *A*, b∈ *B*

**Anti- symmetry:** Let (a_{1}, b_{1}) ≤ (a_{2}, b_{2}) and (a_{2}, b_{2}) ≤ (a_{1}, b_{1})

Then a_{1} ≤ a_{2}, b_{1} ≤ b_{1} and a_{2} ≤ a_{1}, b_{2} ≤ b_{1}

⟹ a_{1} = a_{2}, b_{1} = b_{2}

⟹ (a_{1}, b_{1}) = (a_{2}, b_{2})

**Transitivity:** Let (a_{1}, b_{1}) ≤ (a_{2}, b_{2}) and (a_{2}, b_{2}) ≤ (a_{3}, b_{3})

Then a_{1} ≤ a_{2}, b_{1} ≤ b_{2} and a_{2} ≤ a_{3}, b_{2} ≤ b_{3}

⟹ a_{1} ≤ a_{3} and b_{1} ≤ b_{3}

⟹ (a_{1}, b_{1}) ≤ (a_{3}, b_{3}