**Both the join and meet operations are associative i.e.,**

**a ˅(b˅c) =(a˅b) ˅c and a˄(b˄c)=(a˄b) ˄c**

**Proof:** We first show that, the join operation is associative i.e.,

a ˅(b˅c) =(a˅b) ˅c

Let a ˅(b˅c) = g and (a˅b) ˅c=h

Since g is the least upper bound of a and (b˅c) then a≤g and (b˅c) ≤g

Again, since b≤(b˅c) and c≤(b˅c) then by transitivity

b≤g and c≤g

Now, a≤g and b≤g imply that

(a˅b) ≤g ————————————————-(1)

Again, (a˅b) ≤g and c≤g imply that

(a˅b) ˅c ≤g

⇒ h≤g —————————————————-(2)

Since h is the least upper bound of (a˅b) and c then

(a˅b) ≤h and c≤h

Again, since a≤(a˅b) and b≤ (a˅b) then by transitivity

a≤h and b≤h

Now, b≤h and c≤h imply that

( b˅c) ≤h ———————————————–(3)

Again, a≤h and ( b˅c) ≤h implies that

(a˅b) ˅c ≤h

⇒g ≤h ——————————————–(4)

From (2) and (4) we get , g=h

i.e., a ˅(b˅c) =(a˅b) ˅c

Now by the principle of duality we get,

a˄(b˄c)=(a˄b) ˄c ** (proved)**