Is This Operation an Isomorphism Between These Two Semilattices?

Question

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?

Answer 1

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.

Added. A smaller example:

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

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