Show $x \leq y$ iff $y^c \leq x^c$ in a lattice which is a boolean algebra

Question

Show $x \leq y$ iff $y^c \leq x^c$ where $x,y$ are elements of a lattice $(B,\leq,\lor, \land)$. $(B,\leq)$ is a partially ordered set, and $\lor$ and $\land$ refer to the l.u.b and g.l.b operations respectively.

I started off with the definition of the complement $^c$, which is:

• $x \land x^c = 0$
• $x \lor x^c = 1$

Then, assuming $x \leq y$, I am trying to show $y^c \leq x^c$ - or equivalently, $x^c \leq y^c$ will result in a contradiction. However, I'm unable to proceed. Could someone please help?

Answer 1

If $x \leq y$, we know that $x \land y = x$ and $x \lor y = y$. Also, from De-morgan's laws we know that $(x \land y)^c = x^c \lor y^c$ and $(x \lor y)^c = x^c \land y^c$.

To show $y^c \leq x^c$, we see that:

• $y^c \land x^c = (x\lor y)^c = y^c$
• $y^c \lor x^c = (x \land y)^c = x^c$

Similarly, starting with $y^c \leq x^c$, we can show $x \leq y$, which completes the proof.

Answer 2

What I meant in my comment (before you added that the lattice is a Boolean algebra) is that, for example, with the lattice in the following picture below (where $x'$ is what you denote as $x^c$)

we have $a < b$ but $b^c = d \nleq c = a^c$.
Likewise, $c^c < d^c$ but $d \nleq c$.
So neither of the implications in the equivalence apply.

Of course this is not a Boolean algebra (neither it is a distributive or even modular lattice), and also the de Morgan laws don't apply, since for example, $$(a \wedge b)^c = a^c = c \neq 1 = c \vee d = a^c \vee b^c.$$