3. Main Results
Proposition 4. ( [I; 09]) Let and be residuated lattices.
(i) If and are divisible, then is divisible;
(ii) If and are MTL-algebras and is a chain, then is an MTL-algebra.
Remark 5. 1) If and are BL-algebras and is a chain, then is a BL-algebra.
2) If and are BL-algebras and is not a chain, then is a divisible residuated lattice which is not a BL-algebra, since it is not an MTL-algebra.
In the following, we present some examples of divisible and MTL algebras and an way to obtain its.
Example 6. To generate a divisible residuated lattice with
elements,
which is not an MTL-algebra organized as lattice as in
Figure 1,
We consider the commutative rings and
For the lattice of ideals is , which is a Boolean algebra so a BL-algebra, with the following operations:
Also, we know that is the only MV-chain (up to an isomorphism) with elements, see [FP; 22], so, it is a BL-chain.
The ring has ideals: ..., and
For every
we have
and
We consider two BL-algebras isomorphic with and denoted by and Using Proposition 4, we generate a divisible residuated lattice with elements, for any
This residuated lattice is not an MTL algebra since
For example, for we obtain a divisible residuated lattice(which is not a MTL algebra) with the following operations:
since in
, we have
and
so
is a BL-algebra with 5 elements with the operations:
Examples 7. To generate an MTL-algebra with 8 elements (which is not divisible) organized as a lattice as in
Figure 2,
we consider the MTL-algebra with the following operations:

,
(see [I; 09], p.218) and BL-algebra isomorphic with
Then is an MTL-algebra, by Proposition 4, with the operations:

.
Remark 8. Let Lbe a residuated lattice with 2 elements. Obviously, with

.
Obviously, L is the Boolean algebra thus L has and properties.
Remark 9. Let Lbe a residuated lattice with 3 elements, so Obviously, and L is a chain.
We have the following cases:
1) Obviously, . If a contradiction.
If we obtain a residuated lattice with the following operations:

.
is a BL-algebra isomorphic with , that is not an MV-algebra.
Obviously, satisfies + properties.
If a contradiction.
2) In this case, , so, .
If we obtain a residuated lattice with the following operations:

.
is an MV-algebra isomorphic with , a prime number.
Obviously, satisfies + properties.
If a contradiction.
Remark 10. Let
L be a residuated lattice with 4 elements,
. Then, the lattice
L can be of the form
A or
B from
Figure 3.
Case i.
a and
b are incomparable elements, as in
Figure 3, A.
In this situation, and so, We deduce that so Indeed, if a contradiction.
Analogously,
Moreover, and Indeed, and if then a contradiction.
Also, so since implies a contradiction. It is clear that . Therefore, we obtain a Boolean algebra with the following operations:

izomorphic with
Thus
satisfies
+
properties.
Case ii. The lattice is a chain,
, as in
Figure 3,
B.
We have the following subcases:
1) . We have and . Also, . Therefore, we obtain an MV-algebra structure with the following operations:

.
We remark that is isomorphic with and satisfies + properties.
2) . In this case we do not obtaine a residuated lattice, since ⊙ is not associative. For example, and .
3) . We have and . Also, . Therefore, we obtain a BL-algebra structure (which is not an MV-algebra) with the following tables:
We remark that is isomorphic with and satisfies properties.
4) . We have and . Also, . Therefore, we have and , so, condition is not satisfied. It results that is not a BL-algebra, so it is only an MTL-algebra with the following tables:

.
We remark that is isomorphic with and satisfies property and do not satisfies property.
5) . We have . Also, . Therefore, we have and , false. Condition is not satisfied. It results that is not a BL-algebra. It is only an MTL-algebra with the following tables:

.
We remark that is isomorphic with the residuated lattice see [I; 09], p. 221, and satisfies property and do not satisfies property.
6) , it is not possible, since .
7) . We have . Also, . Therefore, for we obtain a BL-algebra structure (which is not an MV-algebra), isomorphic with with the following tables:

.
We remark that satisfies and properties.
8) is not possible, since .
9) . We have and . Therefore, for we obtain a BL-algebra structure (which is not an MV-algebra) with the following tables:

.
We remark that is isomorphic with and satisfies properties.
10) is not possible, since .
11) is not possible, since .
10) , is not possible, since from we obtain , a contradiction.
Counting the BL-algebras, MTL-algebras and divisible residuated lattices of order 4 we obtain the following result:
Proposition 11.There are 7 residuated lattices of order 4: 2 MV-algebeas 5 BL-algebras, 5 divisible residuated lattices and 7 MTL algebras.
Theorem 12.i) All residuated lattices with elements are MTL-algebras.
ii) There are no divisible residuated lattices with elements that are not MTL-algebras.
In Table 1, we briefly describe the structure of finite residuated lattices L with elements:
|
Nr of residuated lattices |
Structure |
|
1 |
|
|
2 |
|
|
7 |
|
Table 2present a summary for the number of all residuated lattices, Boolean algebras, MV-algebras, BL-algebras, MTL-algebras and divisible residuated lattices with elements.
| |
|
|
|
| All res. latt. |
1 |
2 |
7 |
| Boole |
1 |
- |
1 |
| MV |
1 |
1 |
2 |
| BL |
1 |
2 |
5 |
| DIV |
1 |
2 |
5 |
| MTL |
1 |
2 |
7 |