**Proof:** Let A and B be two lattices the we already proved that A × B = {(a, b) | a ∈ A, b ∈ B} is a poset under the relation ≤ defined by

(a_{1 }, b_{1}) ≤ (a_{2} , b_{2} )⇔ a_{1} ≤ a_{1} in A

b_{1 }≤ b_{2} in B

We show A × B forms a lattice.

Let (a_{1} , b_{1}), (a_{2}, b_{2}) ∈ A × B be any elements.

Then a_{1}, a_{2} ∈ A and b_{1}, b_{2} ∈ B.

### Since A and B are lattices {a_{1}, a_{2}} and {b_{1}, b_{2}} sup and inf in A and B respectively.

Let a_{1} ˄ a_{2} = inf{a_{1}, a_{2}}, b_{1} ˄ b_{2} = inf{b_{1}, b_{2}}

then a_{1} ˄ a_{2} ≤ a_{1}, a_{1} ˄ a_{2} ≤ a_{2}

b_{1} ˄ b_{2} ≤ b_{1}, b_{1} ˄ b_{2} ≤ b_{2}

⟹ (a_{1} ˄ a_{2}, b_{1} ˄ b_{2}) ≤ (a_{1}, b_{1})

(a_{1} ˄ a_{2}, b_{1} ˄ b_{2}) ≤ ( a_{2}, b_{2})

⟹ (a_{1} ˄ a_{2}, b_{1} ˄ b_{2}) is a lower bound of {(a_{1}, b_{1}), ( a_{2}, b_{2})}

*Suppose (c, d) is any lower bound of {(a _{1}, b_{1}), ( a_{2}, b_{2})}*

Then (c, d) ≤ (a_{1}, b_{1})

(c, d) ≤ ( a_{2}, b_{2})

⟹ c ≤ a_{1}, c ≤ a_{2}, d ≤ b_{1}, d ≤ b_{2}

⟹ c is a lower bound of {a_{1}, a_{2}} in A.

D is a lower bound of {b_{1}, b_{2}} in B.

⟹ c ≤ a_{1} ˄ a_{2 }= inf {a_{1}, a_{2}}

d ≤ b_{1} ˄ b2 = inf{b_{1}, b_{2}}

⟹ (c, d) ≤ (a_{1} ˄ a_{2}, b_{1} ˄ b_{2})

i.e., (a_{1} ˄ a_{2}, b_{1} ˄ b_{2}) is g.l.b {(a_{1}, b_{1}), ( a_{2}, b_{2})}

*Similarly by duality we can say that*

(a_{1} ˅a_{2}, b_{1} ˅b_{2}) is l.u.b {(a_{1}, b_{1}), ( a_{2}, b_{2})}

Hence A × B is a lattice.** (Proved)**