**Definition I(according to poset)**: A poset (L, ≤) is said to form a lattice if for every a, b ∈ L, Sup{a, b} and

Inf{a, b} exist in L.

## In that case, we write

Sup{a, b} = a ˅ b

Inf{a, b} = a ˄ b

**Definition II(algebraic):** A non empty set L together with two binary operations ˄ and ˅ is said to form a lattice if ∀ a, b c ∈ L, the following conditions hold:

L1: Idempotency: a ˄ a = a, a ˅ a = a

L2: Commutativity: a ˄ b = b ˄ a, a ˅ b = b ˅ a

L3: Associativity: a ˄ (b ˄ c) = (a ˄ b) ˄ c, a ˅ (b ˅ c) = (a ˅ b) ˅ c

L4: Absorption: a ˄ (a ˅ b) = a, a ˅ (a ˄ b) = a.

**Proof: **By definition, I we find if L is a lattice then for any a, b ∈ L Inf{a, b} = a ˄ b exists and unique. Thus ˄ is a binary operation on L. Similarly ˅ will be a binary operation on L.

Now we have to prove that definition I ⟹ definition II.

Idempotency: a ˄ a = Inf{a, a} = Inf{a} = a

and a ˅ a = Sup{a, a} = Sup{a} = a

Commutativity: a ˄ b = inf {a, b} = inf{b, a} = b ˄ a

and a ˅ b = sup{a, b} = sup{b, a} = b ˅ a

Associativity: Let b ˄ c = d, then d = Inf{b, c}

⟹ d ≤ b, d ≤ c

Let e = Inf{a, d} then e ≤ a, e ≤ d

Thus e ≤ a, e ≤ b, e ≤ c [using transitivity]

⟹ e is a lower bound of {a, b, c}

## If f is any lower bound of {a, b, c} then

f ≤ a, f ≤ b, f ≤ c

Now, f ≤ a, f ≤ b and d = inf{a, b} gives f ≤ d

f ≤ c, f ≤ d and e = inf{c, d} gives f ≤ e

Hence e = inf{a, b, c}

Similarly we can show that (a ˄ b) ˄ c = inf{a, b, c}

### Absorption: Since a ˅ (a ˄ b) is the join of a and (a ˄ b) then

a ˅ (a ˄ b) ≥ a ——————–(i)

Since a ≥ a and a ≥ a ˄ b which implies that

(a ˅ a) ≥ a ˅(a ˄ b)

⟹ a ≥ a ˅(a ˄ b) ——————-(ii)

From (i) and (ii) we get

a ˅ (a ˄ b) = a

Hence from above we can say that definition I ⟹ definition II (Proved)

Now we have to show that definition II⟹ definition I

Let now L be a lattice according to definition II. We first show that L forms a poset. For this we define a relation ≤ on L by a ≤ b ⇔ a ˄ b = a

It is an easy job to show that ≤ so defined is a partial order relation. Thus (L, ≤) is a poset.

Let now a, b ∈ L be any elements, then by definition II, a ˄ b ∈ L. We show a ˄ b = Inf{a, b} with respect to ≤ defined above.

We have (a ˄ b) ˄ a = a ˄ (b ˄ a) By L3

= a ˄ (a ˄ b) By L2

= (a ˄ a) ˄ b

= a ˄ b

⟹ a ˄ b ≤ a

Similarly, a ˄ b ≤ b

⟹ a ˄ b is a lower bound of {a, b}

Suppose c is any lower bound of {a, b} then c ≤ a, c ≤ b

⟹ c ˄ a = c, c ˄ b = c

Thus c ˄ (a ˄ b) = (c ˄ a) ˄ b = c ˄ b = c or that c ≤ a ˄ b is greatest lower bound of {a, b} proving our assertion.

We now show that

a ≤ b ⇔ a ˄ b = a ⇔ a ˅ b = b

Now a ˄ b = a ⟹ (a ˄ b) ˅ b = a ˅ b

⟹ b = a ˅ b by L4

Also a ˅ b = b ⟹ a ˄ (a ˅ b) = a ˄ b

⟹ a = a ˄ b by L4

Hence a ˄ b = a ⇔ a ˅ b = b (⇔ a ≤ b)

Thus by duality we can say that Sup{a, b} will be a ˅ b. Hence definition I hold.** (Proved)**