Extensions and Congruences of Partial Lattices

Definition 2.1.

A partial lattice is a partial algebra (L, ∨, ∧) of type (2,2) satisfying the following identities:

as well as the following duality conditions:

We call weak subalgebras of a partial lattice also partial sublattices.

Lemma 2.1.

Let L be a partial lattice. Then the following hold:

1. L satisfies (x∨y)∧x≈wx and (x∧y)∨x≈wx.

2. If L satisfies (x∨y)∧x≈sx and (x∧y)∨x≈sx, then L is a lattice.

Proof. (i) Let a, b ∈ L and assume (a ∨ b) ∧ a to be defined. Then a ∨ b is defined. Since a∨a=ea we conclude that (a ∨ a) ∨ b is defined. Hence a ∨ (a ∨ b) and (a ∨ b) ∨ a are defined and we obtain

Applying the duality conditions yields (a∨b)∧a=ea. The second weak identity follows analogously.

(ii) is clear.

Example 2.1.

The poset depicted in Figure 1 is not a partially lattice-ordered set since, e.g., U(a,b)={ c,d,1 }≠0, but U(a, b) has no smallest element.

Theorem 2.1.

1. Let L = (L, ∨, ∧) be a partial lattice and ≤ its induced order. Then ℙ(L):=(L,≤) is a partially lattice-ordered set.

2. Let P = (P, ≤) be a partially lattice-ordered set and define partial binary operations ∨ and ∧ on P as follows:

   x∨y{:=sup(x,y)     if  U(x,y)≠0,is  undefinedotherwise, x∧y{:=inf(x,y)     if  L(x,y)≠0,is  undefinedotherwise 

   (x, y ∈ P). Then L(P):=(P,∨,∧) is a partial lattice.

3. Let L = (L, ∨, ∧) be a partial lattice. Then L(ℙ(L))=L.

4. Let P = (P, ≤) be a partially lattice-ordered set. Then ℙ(L(P))=P.

Proof. (i) Let a, b, c ∈ L. Then a∨a=ea=ea∧a and hence a ≤ a. If a ≤ b ≤ a, then a∨b=eb and b∨a=ea and hence a=eb∨a=ea∨b=eb. If a ≤ b ≤ c, then

i.e., a ≤ c. Now assume c ∈ U(a, b). Then a∨c=ec and b∨c=ec and hence

i.e., a ∨ b is defined and a ∨ b ≤ c. Because of

we have a, b ≤ a ∨ b. Together we obtain that in case U(a,b)≠0 the element a ∨ b is defined and coincides with sup(a, b). The dual assertion can be proved analogously.

(ii) Let a, b, c ∈ P. Clearly, a∨a=sa,a∧a=sa,a∨b=sb∨a and a∧b=sb∧a. Assume (a ∨ b) ∨ c to be defined. Since (a ∨ b) ∨ c ∈ U(b, c) the element b ∨ c is defined and since (a ∨ b) ∨ c ∈ U(a, b ∨ c) the element a ∨ (b ∨ c) is defined and we obtain (a ∨ b) ∨ c = sup(a, b, c) = a ∨ (b ∨ c). If, conversely, a ∨ (b ∨ c) is defined, then

The strong identity (x∧y)∧z=sx∧(y∧z) can be proved analogously. Finally, the duality conditions can be easily verified.

(iii) Let ℙ(L)=(L,≤) and L(ℙ(L))=(L,⊔,⊓) and a, b ∈ L. According to the proof of (i), if U(a,b)≠0, then a ∨ b is defined and a ∨ b = sup(a, b). Conversely, if a ∨ b is defined, then

and hence a ∨ b ∈ U(a, b) showing U(a,b)≠0. Hence the following are equivalent: a ∨ b is defined; U(a,b)≠0; a⊔b is defined. In this case we have a⊔b=sup(a,b)=a∨b. Analogously, one can prove that a⊓b is defined if and only if a ∧ b is defined and in this case a⊓b=a∧b.

(iv) If L(P)=(P,∨,∧), ℙ(L(P))=(P,⊑) and a, b ∈ P, then the following are equivalent: a⊑b; a∨b=eb; sup(a.b)=eb; a ≤ b. This shows ℙ(L(P))=P.

Definition 3.1.

Let L = (L, ∨, ∧) be a partial lattice and ≤ its induced order and assume 0,1 ∈ L. Then the two-point extension L^* = (L^*, ≤^*) of L is defined as follows:

1. If ∨ and ∧ are defined everywhere, then

   L∗:=(L,≤).

2. If ∧ is defined everywhere, but ∨ is not, then

   L∗:=(L∪{1},≤∪(L∗×{1})).

3. If ∨ is defined everywhere, but ∧ is not, then

   L∗:=(L∪{0},≤∪({0}×L∗)).

4. If neither ∨ nor ∧ is defined everywhere, then

   L∗:=(L∪{0,1},≤∪({0}×L∗)∪(L∗×{1})).

Clearly, L is a partial sublattice of L*. If L is an infinite lattice having neither a smallest nor a greatest element, then its two-point extension coincides with L. Hence, one should have in mind that L^* neither need have a smallest nor a greatest element.

Lemma 3.1.

Let L = (L, ∨, ∧) be a partial lattice and a, b, c, ∈ L. Then

1. L^* is a lattice with lattice operations

   x∨∗y:=sup≤∗(x,y)andx∧∗y:=inf≤∗(x,y)(x,y∈L∗);

2. a∨b=ecif and only ifa∨∗b=canda∧b=edif and only ifa∧∗b=d;

3. a∨∗b{a∨b    if  U(a,b)≠0,1otherwise      and    a∧∗b{a∧b    if  L(a,b)≠0,0otherwise.

For homomorphisms of partial lattices and their two-point extensions, we can prove the following result.

Theorem 3.1.

Let L_i = (L_i, ∨, ∧), i = 1, 2, be partial lattices.

1. Let h^* be a homomorphism from L1* to L2* with h^*(L₁) ⊆ L₂. Then h^*|L₁ is a homomorphism from L₁ to L₂.

2. Let h be a closed homomorphism from L₁ to L₂. Then there exists some homomorphism h^* from L1* to L2* with h^*|L₁ = h.

Proof.

1. Put h := h^*|L₁ and let a, b ∈ L₁. First assume a ∨ b to be defined. If h(a) ∨ h(b) would not be defined, then we would conclude

   1L2=h(a)∨∗h(b)=h∗(a)∨∗h∗(b)=h∗(a∨∗b)=h(a∨b)∈h(L1)⊆L2,

   a contradiction. Hence h(a) ∨ h(b) is defined and

   h(a∨b)=h∗(a∨∗b)=h∗(a)∨∗h∗(b)=h(a)∨h(b).

   Analogously, one can prove that whenever a ∧ b is defined, also h(a) ∧ h(b) is defined and h(a) ∨ h(b).

2. Define h*:L1*→L2* by

   h∗(x):=h(x)  for all x∈L1,h∗(0L1∗):=0L2∗  if  0L1∗∈L1∗,h∗(1L1∗):=1L2∗  if 1L1∗∈L1∗.

• h^* is well-defined.

  If 0L2*∉L2*, then h(x) ∧ h(y) is defined for all x, y ∈ L₁ and hence, since h is closed, x ∧ y is defined for all x, y ∈ L₁ showing 0L2*∉L2*. Analogously, one can prove that 1L2*∉L2* implies 1L1*∉L1*.

• h^* is a homomorphism from L1* to L2*.

  Let a, b ∈ L₁. If a ∨ b is defined, then h(a) ∨ h(b) is defined and

  h∗(a∨∗b)=h(a∨b)=h(a)∨h(b)=h∗(a)∨∗h∗(b).

  If a ∨ b is not defined then, since h is closed, h(a) ∨ h(b) is not defined, too, and

  h∗(a∨∗b)=h∗(1L1∗)=1L2∗=h(a)∨∗h(b)=h∗(a)∨∗h∗(b).

  Analogously, one can show h^*(a ∧^* b) = h^* (a) ∧* h^*(b). It is not hard to prove

  h∗(x∨∗y)=h∗(x)∨∗h∗(y),h∗(x∧∗y)=h∗(x)∧∗h∗(y)

  for all (x,y)∈(L1*)2\L12.

• h^*|L₁ = h.

  This is clear.

Example 3.1.

The partial lattice L₁ = (L₁, ∨, ∧), visualized in Figure 2, is a partial sublattice of the partial lattice L₂ = (L₂, ∨, ∧) depicted in Figure 3.

Figure 2.

Figure 3.

One can see that id_L1 is a homomorphism from L₁ to L₂ which is not closed. Although L₁ ⊆ L₂, we do not have L1*⊆L2* since L1*≅N5 whereas L2*=L2. Moreover, L1* is not the smallest lattice including L₁ as a partial sublattice.

Definition 4.1.

Let L = (L, ∨, ∧) be a partial lattice. By a congruence on L is meant an equivalence relation E on L satisfying Θ(E) ⋂ L² = E, where Θ(E) denotes the congruence on the lattice L^* generated by E. Let Con L denote the set of all congruences on L. For E ∈ Con L we define two partial operation ∨ and ∧ on L/E by

It is easy to see that the operations ∨ and ∧ on L/E are well-defined.

The next lemma shows that congruences as defined above have the expected properties.

Lemma 4.1.

Let L = (L, ∨, ∧) be a partial lattice. Then

1. Con L = {Θ ⋂ L² | Θ ∈ Con L^*},

2. (Con L, ⊆) is a complete lattice.

Proof.

1. Let E be an equivalence relation on L. If E ∈ Con L, then Θ(E) ∈ Con L^* and E = Θ(E)∩L². If, conversely, Θ ∈ Con L^* and Θ ⋂ L² = E, then E ⊆ Θ and hence Θ(E) ⊆ Θ whence

   E⊆Θ(E)∩L2⊆Θ∩L2=E

   which implies Θ(E) ∩ L² = E and hence E ∈ Con L.

2. If E_i ∈ Con L for all i ∈ I, then for every i ∈ I, there exists some Θ_i ∈ Con L^* satisfying Θ_i ∩ L² = E_i and hence ∩i∈IΘi∈Con L* and

   ∩i∈IEi=∩i∈I(Θi∩L2)=(∩i∈IΘi)∩​L2∈  Con  L

   according to (i).

Lemma 4.2.

Let L = (L, ∨, ∧) be a partial lattice, a, b ∈ L and E ∈ Con L. Then

where α ∈ [1](Θ(E)) \ {1}.

Proof. If U(a,b)≠0, then

and hence [a] E ∨ [b] E = [a ∨ b] E. If U(a,b)=0 and [1](Θ(E)) = {1}, then

and hence [a] E ∨ [b] E is not defined. Assume, finally, U(a,b)=0 and [1](Θ(E)) ≠ {1}. then [ 1 ](Θ(E))∩L=0 would imply [1](Θ(E)) = {0,1} and hence Θ(E) = (L^*)² whence L=L*∩L=[ 1 ](Θ(E))∩L=0, a contradiction. This shows that there exists some α ∈ [1](Θ(E)) ∩ L. Now

and hence [a]E ∨ [b]E = [α]E.

Remark 4.1.

Due to duality, the dual version of Lemma 4.2 holds, too.

The main point is to show that the quotient partial algebra (L/E, ∨, ∧) as defined above is again a partial lattice, the so-called quotient partial lattice by the congruence E.

Theorem 4.1.

Let L = (L, ∨, ∧) be a partial lattice and E ∈ Con L. Then L/E := (L/E, ∨, ∧) is a partial lattice.

Proof. In this proof we often use Lemma 4.2. We prove only a part of the conditions mentioned in Definition 2.1, the remaining conditions follow by duality. Let a, b, c ∈ L. It is evident directly by Definition 4.1 that the partial operations ∨ and ∧ on L/E satisfy strong idempotency and strong commutativity. We need to prove also strong associativity. Assume ([a]E ∨ [b]E) ∨ [c]E to be defined. We have

and in this case we conclude

• Case 1. U(a, b), U(a∨b,c)≠0.

  Then [a]E ∨ ([b]E ∨ [c]E) is defined and

  ([ a ]E∨[ b ]E)∨[ c ]E=[ a∨b ]E∨[ c ]E=[ (a∨b)∨c ]E=[ a∨(b∨c) ]E=[ a ]E∨[ b∨c ]E=[ a ]E∨([ b ]E∨[ c ]E).

• Case 2. U(a,b)=0 or ( U(a,b)=0 and U (a∨b,c)=0 ).

  Then 1 ∈ L^*, [1](Θ(E)) = {1} and [a]E ∨ ([b]E ∨ [c]E) is defined. According to the proof of Lemma 4.2, there exists some α ∈ [1](Θ(E)) ∩ L and

  [ x ]E∨[ α ]E=[ x∨*α ](Θ(E))∩L=[ x∨*1 ](Θ(E))∩L=[ 1 ](Θ(E))∩L=[ α ](Θ(E))∩L=[ α ]Efor all x∈L.

• Case 2a. U(a,b)=U(b,c)=0.

  Then

  ([ a ]E∨[ b ]E)∨[ c ]E=[ α ]E∨[ c ]E=[ α ]E=[ a ]E∨[ α ]E=[ a ]E∨([ b ]E∨[ c ]E).

  Hence ([a]E ∨ [b]E) ∨ [c]E = [a]E ∨ ([b]E ∨ [c]E).

• Case 2b. U(a,b)=0 and U(b,c)≠0.

  Then U(a,b∨c)=0 and

  ([ a ]E∨[ b ]E)∨[ c ]E=[ α ]E∨[ c ]E=[ α ]E=[ a ]E∨[ b∨c ]E=[ a ]E∨([ b ]E∨[ c ]E).

• Case 2c. U(a,b)=0 and U(b,c)=0.

  Then U(a∨b,c)=0 and

  ([ a ]E∨[ b ]E)∨[ c ]E=[ a∨b ]E∨[ c ]E=[ α ]E=[ a ]E∨[ α ]E=[ a ]E∨([ b ]E∨[ c ]E).

• Case 2d. U(a,b),U(b,c)=0 and U(a∨b,c)=U(a,b∨c)=0.

  Then

  ([ a ]E∨[ b ]E)∨[ c ]E=[ a∨b ]E∨[ c ]E=[ α ]E=[ a ]E∨[ b∨c ]E=[ a ]E∨([ b ]E∨[ c ]E).

  If, conversely, [a]E ∨ ([b]E ∨ [c]E) is defined, then

  [ a ]E∨([ b ]E∨[ c ]E)=e([ b ]E∨[ c ]E)∨[ a ]E=e([ c ]E∨[ b ]E)∨[ a ]E=e[ c ]E∨([ b ]E∨[ a ]E)=e([ b ]E∨[ a ]E)∨[ c ]E=e([ a ]E∨[ b ]E)∨[ c ]E.

  Finally, the duality conditions remain to be proved. If [ a ]E∨[ b ]E=e[ a ]E, then a ∈ [a]E ⊆ [a ∨^* b] (Θ(E)), i.e. a Θ(E) (a∨*b) and hence (a ∧^* b) Θ(E) ((a ∨^* b) ∧^* b) = b whence [ a∧*b ](Θ(E))∩L=[ b ](Θ(E))∩L=[ b ]E≠0, i.e. [ a ]E∧[ b ]E=e[ b ]E.

Lemma 4.3.

Let L = (L, ∨, ∧) be a partial lattice and E ∈ Con L and define h: L → L/E by h(x) := [x]E for all x ∈ L. Then h is a homomorphism from L to L/E and ker h = E. If, in addition, [0_L^*](Θ(E)) = (0_L^*} provided 0_L^* ∈ L^* and [1_L^*](Θ(E)) = {1_L^*} provided 1_L^* ∈ L^*, then h is a closed homomorphism from L to L/E.

Proof. This easily follows from Lemma 4.2.

Theorem 4.2.

For i = 1, 2 let L_i = (L_i, ∨, ∧) be partial lattices and, h a closed homomorphism from L₁ to L₂. Then ker h ∈ Con L₁. If, in addition, [ 0L1* ](Θ(kerh))={ 0L1* } provided 0L1*∈L1* and [ 1L1* ] (Θ(kerh))={ 1L1* } provided 1L1*∈L1*, then (h(L₁), ∨, ∧) (with partial operations defined as in L₂) is a partial lattice which is isomorphic to L₁/(ker h).

Proof. According to Theorem 3.1, there exists some homomorphism h^* from L1* to L2* satisfying h^*|L₁ = h. Now ker h^* ∈ Con L₁ and hence ker h=ker h*∩L12∈Con L1. If the additional condition holds, then, according to Lemma 4.2, the mapping h(x) ↦ [x] ker h is a well-defined isomorphism from (h(L₁), ∨, ∧) to L₁/(ker h) since for a, b ∈ L₁ the following are equivalent: h(a) ∨ h(b) exists in L₂; a ∨ b exists in L₁; [a] ker h ∨ [b] ker h exists in L₁/(ker h). Analogous statements hold for ∧ instead of ∨.

Theorem 4.3.

Let L = (L, ∨, ∧) be a partial lattice and E ∈ Con L. Then (L/E)^* ≅ L^*/(Θ(E)).

Proof. Define f: (L/E)^* → L^*/(Θ(E)) in the following way:

We will prove that f is an isomorphism from (L/E)^* to L^*/(Θ(E)). Let a, b ∈ L.

• f is well-defined.

  If [a]E = [b]E, then [a] (Θ(E)) = [b](Θ(E)). We prove that U([ a ]E,[ b ]E)=0 implies U(a,b)=0. Assume U(a,b)=0, say c ∈ U(a, b). Then a∨c=ec and b∨c=ec. Hence

  [ a ]E∨[ c ]E=[ a∨*c ](Θ(E))∩L=[ c ](Θ(E))∩L=[ c ]E

  and, analogously, [b]E ∨ [c]E = [c]E, i.e. [ c ]E∈U([ a ]E,[ b ]E)≠0. Hence, if 1_(L/E)^* ∈ (L/E)^*, then there exist d, e ∈ L with U([ d ]E,[ e ]E)=0. Therefore U(d,e)=0 which shows 1_L^* ∈ L^*. Analogously, one can prove that 0_(L/E)^* ∈ (L/E)^* implies 0_L^* ∈ L^*.

• f is injective.

  If [a](Θ(E)) = [b](Θ(E)), then (a, b) ∈ Θ(E) ∩ L² = E, i.e. [a]E = [b]E. Now assume 1_(L/E)^* ∉ (L/E)^* and [a](Θ(E)) = [1_L^*](Θ(E)). Then there exist c, d ∈ L with U([ c ]E,[ d ]E)=0. Now we have

  [ c ]E∨[ a ]E=[ c∨*a ](Θ(E))∩L=[ c∨*1L* ](Θ(E))∩L=[ 1L* ](Θ(E))∩L=[ a ](Θ(E))∩L=[ a ]E

  and, analogously, [d]E ∨ [a]E = [a]E. This shows [a]E ∈ U([c]E, [d]E), a contradiction. Hence [a](Θ(E)) ≠ [1_L^*](Θ(E)). Analogously, one obtains that in case 0_(L/E)^* ∈ (L/E)^* we have [a](Θ(E)) ≠ [0_L^*](Θ(E)). Now assume 0_(L/E)^*, 1_(L/E)^* ∈ (L/E)^* and [0_L^*](Θ(E)) = [1_L^*](Θ(E)). Then 0_L^*, 1_L^* ∈ L^* and (0_L^*, 1_L^*) ∈ Θ(E) and hence Θ(E) = (L^*)² whence

  E=Θ(E)∩L2=(L*)2∩L2=L2

  which implies |L/E| = 1 and hence 0_(L/E)^*, 1_(L/e)^* ∉ (L/E)^*, a contradiction.

• f is surjective.

  Assume 1_L^* ∈ L^*. If 1_(L/E)^* ∈ (L/E)^*, then f(1_(L/E)^*) = [1_L^*](Θ(E)). Now assume 1_(L/E)^* ∉ (L/E)^*. Then [x]E∨[y]E is defined in L/E for every x, y ∈ L. Suppose [1_L^*](Θ(E)) = {1_L^*}. Then U(x,y)≠0 for all x,y ∈ L and hence 1_L^* ∉ L^*, a contradiction. Therefore [1_L^*](Θ(E)) ≠ {1_L^*}. According to the proof of Lemma 4.2, there exists some α ∈ [1_L^*](Θ(E)) ∩ L and we obtain f([α]E) = [α](Θ(E)) = [1_l^*](Θ(E)). Analogously, one can show that in case 0_L^* ∈ L^*, there exists some x ∈ (L/E)^* with f(x) = [0_L^*](Θ(E)).

• f is a homomorphism from (L/E)^* to L^*/(Θ(E)).

  First assume U(a,b)≠0. Then

  f([ a ]E∨[ b ]E)=f([ a∨*b ](Θ(E))∩L)=f([ a∨b ](Θ(E))∩L)=f([ a∨b ]E)=[ a∨b ](Θ(E))=[ a ](Θ(E))∨[ b ](Θ(E))=f([ a ]E)∨f([ b ]E).

  Now assume U(a,b)=0. Then 1_L^* ∈ L^*. First assume [1_L^*](Θ(E)) = {1_L^*}. Then 1_(L/E)^* ∈ (L/E)^* and

  f([ a ]E∨[ b ]E)=f(1(L/E)*)=[ 1L* ](Θ(E))=[ a∨b ](Θ(E))=[ a ](Θ(E))∨[ b ](Θ(E))=f([ a ]E)∨f([ b ]E).

  Finally, assume [1_L^*](Θ(E)) ≠ {1_L^*}. According to the proof of Lemma 4.2, there exists some α ∈ [1_L^*](Θ(E)) ∩ L, and we have

  f([ a ]E∨[ b ]E)=f([ a∨*b ](Θ(E))∩L)=f([ 1L* ](Θ(E))∩L)=f([ α ](Θ(E))∩L)=f([ α ]E)=[ α ](Θ(E))=[ 1L* ](Θ(E))=[ a∨b ](Θ(E))=[ a ](Θ(E))∨[ b ](Θ(E))=f([ a ]E)∨f([ a ]E).

  It is easy to see that f(x ∨ y) = f(x) ∨ f(y) for all (x,y) ∈ ((L/E)^*)² \ (L/E)². Analogously, one can prove f(x ∧ y) = f(x) ∧ f(y) for all x, y ∈ (L/E)^*.

Example 5.1.

Let L denote the partial lattice visualized in Figure 4.

Figure 4.

The lattice L^* is depicted in Figure 5.

Figure 5.

Here 0_L^*, 1_L^* ∈ L^*. Put E := {a, c}² ∪ {b}² Then Θ(E) = {0_L^* }² ∪ {a, c}² ∪ {b}² ∪ {1_L^*}² and hence Θ(E) is a congruence on L^* satisfying Θ(E) ∩ L² = E and L/E is visualized in Figure 6.

Figure 6.

The lattice (L/E)^* is depicted in Figure 7,

Figure 7.

and we have 0_(L/e)^*, 1_(L/e)^* ∈ (L/E)^*.

Finally, L^*/(Θ(E)) is visualized in Figure 8.

Figure 8.

In accordance with Theorem 4.3, (L/E)^* ≅ L^*/(Θ(E)).

Example 5.2.

Let L be the partial lattice depicted in Figure 9.

Figure 9.

The lattice L^* is visualized in Figure 10.

Figure 10.

Here 0_L^*, 1_L^* ∈ L^*. Put E := {a}² ∪ {b, d}² ∪ {c}². Let us mention that (b, d) ∈ E and a ∨ b = c, but a ∨ d does not exist in L. We have Θ(E) = {0_L^*}² ∪ {a}² ∪ {b, d}² ∪ {c, 1_L^*}² and hence Θ(E) ∩ L² = E and L/E is depicted in Figure 11.

Figure 11.

One can see that although [a]E ∨ [d]E exists in L/E, a ∨ d does not exist in L. The lattice (L/E)^* is visualized in Figure 12,

Figure 12.

and we have 0_(L/E)^* ∈ (L/E)^* and 1_(L/E)^* ∉ (L/E)^*.

Finally, L^*/(Θ(E)) is depicted in Figure 13.

Figure 13.

Example 5.3.

Let L denote the partial lattice from Example 5.2. Then 0_L^*, 1_L^* ∈ L^*. Put E := {a, c}² ∪ {b}² ∪ {d}². Here (a, c) ∈ E and c ∧ d = b, but a ∧ d does not exist in L. We have Θ(E) = {0_L^*, b}² ∪ {a, c}² ∪ {d}² ∪ {1_L^* }² and hence Θ(E) ∩ L² = E and L/E is visualized in Figure 14.

Figure 14.

One can see that although [a]E∧ [b]E exists in L/E, a∧b does not exists in L. The lattice (L/E)^* is depicted in Figure 15.

Figure 15.

and we have 0_(L/E)^* ∉ (L/E)^* and 1_(L/E)^* ∈ (L/E)^*.

Finally, L^*/(Θ(E)) is visualized in Figure 16.

Figure 16.

The next example shows that L/E may be a lattice even if L is only a partial lattice.

Example 5.4.

Let L denote the partial lattice from Example 5.2. Then 0_L^*, 1_L^* ∈ L^*. Put E := {a}² ∪ {b, c}² ∪ {d}². Then Θ(E) = {0_L^*, a}² ∪ {b, c}² ∪ {d, 1_l^*L² and hence Θ(E) ∩ L² = E and L/E is depicted in Figure 17.

Figure 17.

Hence (L/E)^* = L/E and 0_(L/E)^*, 1_(L/E)^* ∉ (L/E)^*.

Finally, L^*/(Θ(E)) is visualized in Figure 18.

Figure 18.