# Operation table of Hasse diagram

Consider the following Hasse diagram:

My book gives the following join and meet operation tables for this diagram:

$$\begin{array}{|c || c | c|} \hline Subset & x \wedge y & x \vee y \\ \hline \{a,b\} & 0 & 1 \\ \{a,c\} & c & a \\ \{a,1\} & a & 1 \\ \{a,0\} & 0 & a \\ \{b,c\} & 0 & 1 \\ \{b,a\} & 0 & 1 \\ \{b,1\} & b & 1 \\ \{b,0\} & 0 & b \\ \{c,a\} & c & a \\ \{c,1\} & c & 1 \\ \{c,b\} & b & 1 \\ \{c,0\} & 0 & c \\ \hline \end{array}$$

I don't follow, why,

$a\land c$ = $c$ and $a\lor c$ =$a$, shouldn't it be vice versa ?

$a\land b$ = $0$ ?

$a\lor b$ = $1$ ?

It would be nice if someone could help me out here.

• It seems to me that $a\wedge c=a$ and $a\vee c=c$ as you suggested and $c\wedge b=0$. Off course $\left\{ b,c\right\} =\left\{ c,b\right\}$ but the outcomes $c\wedge b$ and $b\wedge c$ are different. Quite some things are wrong here. $a\wedge b=0$ and $a\vee b=1$ looks correct to me ( I am not an expert in this). – drhab Apr 30 '14 at 17:10

$$\begin{array}{|c || c | c|} \hline \text{Subset} & x \wedge y & x \vee y \\ \hline \{a,b\} & 0 & 1 \\ \{a,c\} & a & c \\ \{a,1\} & a & 1 \\ \{a,0\} & 0 & a \\ \{b,c\} & 0 & 1 \\ \{b,1\} & b & 1 \\ \{b,0\} & 0 & b \\ \{c,1\} & c & 1 \\ \{c,0\} & 0 & c \\ \hline \end{array}$$
Now why do we have $a \wedge b = 0$ and $a \vee b = 1$? In terms of the Hasse diagram, $0$ is the first element "downwards accessible" from both $a$ and $b$; likewise for $1$, but "upwards".