Proving Complemented Distributive Lattice Property

[namya]

Show that in a complemented distributive lattice a ≤ b ⇔ a ∗ bʹ = 0 ⇔ aʹ ⊕ b = 1 ⇔ aʹ ≤ bʹ.
can somebody help me prove this.

 

[Valkarie]

A:We will prove that $a\le b$ iff $a'\le b'$. We'll do this by proving the two implications separately.$\underline{\text{If }a\le b,\text{ then }a'\le b'}$Let $a,b$ be elements of a complemented distributive lattice. Assume that $a\le b$. Then, since $a\le b$, it follows that $a\land b=a$.Now, consider $a'\lor b'$. Since $a\land b=a$, we have$$a'\lor b'=(a\land b)'=a'\lor b$$by De Morgan's Law. But since $a\le b$, this implies that $a'\le b'$ by definition of complemented distributive lattices. $\underline{\text{If }a'\le b',\text{ then }a\le b}$Let $a,b$ be elements of a complemented distributive lattice. Assume that $a'\le b'$. Then, since $a'\le b'$, it follows that $a'\lor b'=b'$.Now, consider $a\land b$. Since $a'\lor b'=b'$, we have$$a\land b=(a\lor b')'=b''=b$$by De Morgan's Law. But since $a'\le b'$, this implies that $a\le b$ by definition of complemented distributive lattices.Therefore, we have shown that $a\le b$ iff $a'\le b'$. $\blacksquare$