Is This Operation an Isomorphism Between These Two Semilattices?

If $(L, \land, \lor, 0, 1)$ is a lattice, and there exists a unary operation $'$ on $L$ such that

1. $(x \lor x')=1$, and

2. $(x \land x')=0$

both hold, is the unary operation $'$ an isomorphism between the semilattices $(L, \land)$ and $(L, \lor)$? If we add the condition that $'$ is an involution (i. e. $x''=x$), is $'$ an isomorphism?

-

No.

Consider this lattice:

                                    1
/ \
/   \
/     \
d       w
/ \     / \
b   c   y   z
\ /     \ /
a       x
\     /
\   /
\ /
0


and make $'$ exchange $0\leftrightarrow 1$, $d\leftrightarrow w$, $b\leftrightarrow y$, $c\leftrightarrow z$, $a\leftrightarrow x$. It is easy to verify that $r\land r' = 0$ and $r\lor r' = 1$ for all $r$; but $'$ does not define an isomorphism from $(L,\land)$ to $(L,\lor)$, since for example $b'\lor c' = y\lor z = w$, but $(b\land c)' = a' = x\neq w$. Nor does it define an isomorphism going the other way, since the map is self-invertible.

                                     1
/ \
/   \
b     y
|     |
a     x
\   /
\ /
0


But $a'\lor b' = x\lor y = y$, $(a\land b)' = a' = x$.

-
Did you use any particular method to find this? –  Doug Spoonwood Feb 16 '12 at 5:05
It suggested itself fairly quickly: take a disjoint union of two copies of the same lattice, stick a $0$ and a $1$ to that, then have $'$ swap the copies. Have enough elements to have a nontrivial join and a nontrivial meet, and make it as small as possible. Had I thought a bit more, I could have made it even smaller: remove $d$ and $w$, or remove $a$ and $x$; you don't need both a nontrivial join and a nontrivial meet, just one of them. –  Arturo Magidin Feb 16 '12 at 5:09
Or even smaller: remove $d$, $w$, $c$, and $z$. –  Arturo Magidin Feb 16 '12 at 5:13
I don't know what you mean by "nontrivial join (meet)". If we have an unordered pair {x, y} such that (x $\lor$ y)=z, where z $\neq$ y and z $\neq$ x, would you call their meet nontrivial? Or is there more to it? –  Doug Spoonwood Feb 16 '12 at 13:24
I meant "Nontrivial meet"$\neq 0$, "nontrivial join$\neq 1$. The second example shows you don't need $x\lor y\notin\{x,y\}$ for a counterexample to work. –  Arturo Magidin Feb 16 '12 at 16:05