**Proof:** Let *A* and *B* be the given self dual posets.

Let *f*: *A*→* A* and g:

*B*→

*be the isomorphisms.*

__B__Now we define *h*: *A × B* → *A × B*, such that,

*h*((a, b)) = (*f*(a), g(b))

then h is well defined, one – one map as

(a_{1}, b_{1}) = (a_{2}, b_{2})

⇔ a_{1} = a_{2}, b_{1} = b_{2}

⇔ *f *(a_{1}) = *f* (a_{2}), *g* (b_{1}) =* g* (b_{2})

⇔ ( *f *(a_{1}, *g*(b_{1})) = ( *f *(a_{2}), *g* (b_{2}))

⇔ h((a_{1}, b_{1})) = h((a_{2}, b_{2}))

Again for any (x, y) ∈ __A × B__, since

h (*f ^{ -1}*(x),

*g*(y)) = (

^{–1}*ff*(x),

^{–1}*gg*(y)) = (x, y)

^{ –1}and *f ^{-1}, g ^{-1}* exist as

*f, g*are one-one onto, we find h is onto.

Again (a_{1}, b_{1}) ≤ (a_{2}, b_{2})

⇔ a_{1} ≤ a_{2}, b_{1} ≤ b_{2}

⇔ *f *(a_{1}) ≥ *f* (a_{2}), *g* (b_{1}) ≥* g* (b_{2})

⇔ ( *f *(a_{1}, *g*(b_{1})) ≥ ( *f *(a_{2}), *g* (b_{2}))

⇔ *h*((a_{1}, b_{1})) ≥ *h*((a_{2}, b_{2}))

*Thus h is the required isomorphism and A × B ≅ A × B. (Proved)*