Construction of Quasi-Modular Double 𝑆-Algebras
S. El Assar, M. Atallah and E. Atef
Math. Dept., Faculty of Science, Tanta Univ., Egypt.
Abstract:
In this paper we characterize and construct a special class of
quasi-modular double p-algebras which satisfy the Stone identity,
using a quadruple. The basic idea is to associate every such
algebra with two simpler structures and two connecting maps.
We describe, in terms of quadruples, homomorphisms and
subalgebras of quasi-modular double 𝑆-algebras. We also,
characterize the congruence relations of quasi-modular double 𝑆algebras.
1: Introduction
T.katrinak ([9]), [7] characterized (quasi)-modular p-algebras in
terms of Boolean algebras, lattices with 1 and certain connecting
maps. Since every quasi-modular double p-algebra is a special
quasi- modular double p-algebra, it makes sense that a similar
method of construction of of quasi-modular double p-algebras from
simpler structures can be found out.
In [4], T.katrinak gave such a method of construction for regular
double p-algebras. In his paper [5], he evolved such a method of
construction of modular double S-algebras.
We show in the present paper, that a similar method can be
extended to quasi-modular double S-algebras which form a larger
class of the modular one.
-1-
Every quasi-modular double S-algebra is associated with two
simpler structures and two connective maps, forming a quadruple.
We prove the correspondence between quasi-modular double Salgebras and quadruples.
We characterize, in terms of quadruples, homomorphisms,
subalgebras and congruence relations of the algebra involved.
Some studies on the class of quasi-modular double p-algebras
are given in [2], [6].
2: Preliminaries:
An algebra< 𝐿;∧,∨ , * , 0,1 > of type < 2,2,1,0,0 > is called a
p-algebra (pseudocomplementd lattice or PCL) if < 𝐿;∧,∨ ,0,1 > is a
bounded lattice and for every 𝑎 ∈ 𝐿, * is a unary operation on 𝐿
such that
𝑎∧𝑥 =0
𝑖𝑓𝑓 𝑥 ≤ 𝑎∗ ,
An algebra < 𝐿;∧,∨ , * , +, 0,1 > is called a double p-algebra if <
𝐿,∧,∨ , * , 0, 1 > and < 𝐿,∧,∨, +,0,1 > is a dual p-algebra i.e
𝑎∨𝑥 =1
𝑖𝑓𝑓 𝑥 ≥ 𝑎+ ,
For the standard rules of computation in (double) p-algebras,
we refer to [3],[4] and [10]. By a distributive (modular) p-algebra <
𝐿;∧,∨ , * , 0,1 > we mean that the lattice 𝐿 is distributive (modular).
Further, we recall that a Stone algebra is a distributive p-algebra
satisfying the Stone identity
𝑎∗ ∨ 𝑎∗∗ = 1
-2-
In general p-algebras satisfying the Stone identity are called
𝑆-algebras. A larger variety of quasi-modular p-algebras was
introduced by T.Katrinak and Mederly [8]. This subvarity of palgebras is defined by the identity
((𝑥 ∧ 𝑦) ∨ 𝑧 ∗∗ ) ∧ 𝑥 = (𝑥 ∧ 𝑦) ∨ (𝑥 ∧ 𝑧 ∗∗ )
(1)
The quasi-modular p-algebras satisfy the identity
𝑥 = 𝑥 ∗∗ ∧ (𝑥 ∨ 𝑥 ∗ ) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐿
(2)
An algebra < 𝐿;∧,∨ , * , +, 0,1 > is called a quasi-modular
double p-algebra if < 𝐿;∧,∨ , * , 0,1 > is a quasi-modular p-algebra
and < 𝐿;∧,∨, +, 0, 1 > is a dual quasi-modular p-algebra.
A double 𝑆-algebra is a double p-algebra satisfying the identities
𝑎∗ ∨ 𝑎∗∗ = 1 𝑎𝑛𝑑 𝑎+ ∧ 𝑎++ = 0
A double Stone algebra is a distributive double 𝑆-algebra .
In a double p-algebra 𝐿, the sets 𝐵(𝐿) = {𝑥 ∈ 𝐿: 𝑥 = 𝑎∗∗ } and
̅̅̅̅̅̅ = {𝑥 ∈ 𝐿: 𝑥 = 𝑎++ } give Boolean algebras < 𝐵(𝐿),∧,⊔ , *
𝐵(𝐿)
̅̅̅̅̅̅,⊓,∨, +, 0, 1 > where 𝑎 ⊔ 𝑏 = (𝑎∗ ∧ 𝑏 ∗ )∗ and 𝑎 ⊓
, 0,1 > and < 𝐵(𝐿)
𝑏 = (𝑎+ ∨ 𝑏 + )+ .
If 𝐿 is a a quasi-modular double 𝑆- algebra, then 𝐵(𝐿)(=
̅̅̅̅̅̅) is a subalgebra of 𝐿 and 𝑥 ∗+ = 𝑥 ∗∗ , 𝑥 +∗ = 𝑥 ++ .
𝐵(𝐿)
̅̅̅̅̅̅̅ = {𝑥 ∈ 𝐿: 𝑥 + = 1}
The sets 𝐷(𝐿) = {𝑥 ∈ 𝐿: 𝑥 ∗ = 0} and 𝐷(𝐿)
are a filter and an ideal of 𝐿, respectively.
Congruences on double p-algebras are lattice congruences
preserving the operations * , + . The congruence relation Φ of a
double p-algebra defined by
-3-
𝑥 ≡ 𝑦(Φ) 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 ∗ = 𝑦 ∗ 𝑎𝑛𝑑
𝑥+ = 𝑦+
Is called the determination congruence and a member of 𝐿⁄Φ is
referred to as a determination class of 𝐿. In a double 𝑆-algebra we
̅̅̅̅̅̅ , and the lattice of congruence relations
have 𝐿⁄Φ ≅ 𝐵(𝐿) = 𝐵(𝐿)
of 𝐿will be denoted by 𝐶𝑜𝑛(𝐿).
For the determination congruence Φ we define for all 𝑎 ∈
𝐵(𝐿), the set 𝐹𝑎 = {𝑥 ∈ 𝐿: 𝑥 ∗∗ = 𝑎} makes a partition of 𝐿 and
so 𝐹̅𝑎 = {𝑥 ∈ 𝐿: 𝑥 ++ = 𝑎} .In a double 𝑆-algebra 𝐹𝑎 = 𝐹̅𝑎 and 𝐹𝑎 ≅
𝐿 ⁄Φ
The following rules of computations will be used frequently,
see [1], [3].
In p-algebras the following rules are true:
(1)𝑎 ≤ 𝑏 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑏 ∗ ≤ 𝑎∗ ,
(2)𝑎 ≤ 𝑎∗∗ ,
(3)𝑎∗ = 𝑎∗∗∗ ,
(4)(𝑎 ∨ 𝑏)∗ = 𝑎∗ ∧ 𝑏 ∗ ,
(5)(𝑎 ∧ 𝑏)∗∗ = 𝑎∗∗ ∧ 𝑏 ∗∗ ,
(6)(𝑎 ∧ 𝑏)∗ ≥ 𝑎∗ ∨ 𝑏 ∗ ,
(7)(𝑎 ∨ 𝑏)∗∗ = (𝑎∗∗ ∨ 𝑏 ∗∗ )∗∗ ≥ 𝑎∗∗ ∧ 𝑏 ∗∗ .
In 𝑆-algebra
(8)(𝑎 ∧ 𝑏)∗ = 𝑎∗ ∨ 𝑏 ∗ ,
(9)(𝑎 ∨ 𝑏)∗∗ = 𝑎∗∗ ∧ 𝑏 ∗∗ .
-4-
In a double p-algebra, we can formulate the above identities
and inequalities in the dual p-algebra and we can prove the
following
(10) 𝑎+ ≤ 𝑎∗ ;
(11)𝑎+∗ ≤ 𝑎++ ≤ 𝑎 ≤ 𝑎∗∗ ≤ 𝑎∗+ ;
(12)(𝑎 ∧ 𝑏)+∗ = 𝑎+∗ ∧ 𝑏 +∗ .
In double S-algebra
(13)𝑎∗+ = 𝑎∗∗ ,
𝑎+∗ = 𝑎++ ;
(14)(𝑎 ∧ 𝑏)∗ = 𝑎∗ ∨ 𝑏 ∗ 𝑎𝑛𝑑 (𝑎 ∨ 𝑏)+ = 𝑎+ ∧ 𝑏 + ;
(15)(𝑎 ∨ 𝑏)∗∗ = 𝑎∗∗ ∧ 𝑏 ∗∗ 𝑎𝑛𝑑 (𝑎 ∧ 𝑏)++ = 𝑎++ ∨ 𝑏 ++ .
Lemma 1:
Let 𝐿 be a double 𝑆-algebra, then
𝑥 ∗∗+ = 𝑥 ∗ 𝑎𝑛𝑑 𝑥 ++∗ = 𝑥 +
Proof:
Since, 𝑥 + ∧ 𝑥 ++ = 0 then 𝑥 + ≤ 𝑥 ++∗ holds, and from the
identity (2), put 𝑎 = 𝑥 + ,𝑥 ++∗ ≤ 𝑥 + . Hence, 𝑥 + = 𝑥 ++∗ .
Similarly, 𝑥 ∗ ∨ 𝑥 ∗∗ = 1 then 𝑥 ∗ ≥ 𝑥 ∗∗+ holds. But , 𝑎 ≤ 𝑎∗+ ,
put 𝑎 = 𝑥 ∗ ,𝑥 ∗ ≤ 𝑥 ∗∗+ . Hence, 𝑥 ∗ = 𝑥 ∗∗+ .
3: Characterization of Quasi-Modular Double 𝑺-Algebra
We recall that a double p-algebra is called a quasi-modular
double p-algebra if the identities
((𝑥 ∧ 𝑦) ∨ 𝑧 ∗∗ ) ∧ 𝑥 = (𝑥 ∧ 𝑦) ∨ (𝑥 ∧ 𝑧 ∗∗ )
-5-
(1)
And ((𝑥 ∨ 𝑦) ∧ 𝑧 ++ ) ∨ 𝑥 = (𝑥 ∨ 𝑦) ∧ (𝑥 ∨ 𝑧 ++ )
(2)
are satisfied in 𝐿.
T. Katrinak and Mederly [8] proved that, the class of quasimodular p-algebras satisfied the identity
𝑥 = 𝑥 ∗∗ ∧ (𝑥 ∨ 𝑥 ∗ ) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐿
(3)
Also, the class of quasi-modular dual p-algebras satisfies the
identity
𝑥 = 𝑥 ++ ∨ (𝑥 ∧ 𝑥 + )
𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐿
(4)
Hence, the quasi-modular double p-algebras satisfies the
identities (3), (4).
Let 𝐿 be a quasi-modular p-algebra. The set 𝐵(𝐿) is a
Boolean algebra, 𝐷(𝐿) is a filter of 𝐿. 𝐷(𝐿) is a lattice with the
largest element 1. Denote by 𝐹(𝐷(𝐿)) the set of of all filters of
𝐷(𝐿). Consider the map 𝜑(𝐿): 𝐵(𝐿) ⟶ 𝐹(𝐷(𝐿)) defined by
(𝑎)𝜑(𝐿) = [𝑎∗ ) ∩ 𝐷(𝐿), 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎 ∈ 𝐵(𝐿).
Theorem 1 [8, Theorem 3 in 6.2].
Let 𝐿 be a quasi-modular p-algebra. Then the structure map
𝜑(𝐿): 𝐵(𝐿) ⟶ 𝐹(𝐷(𝐿)) is a {0,1,∨}-homomorphism. In addition, for
every 𝑎 ∈ 𝐵(𝐿) we have
𝑎𝜑(𝐿) ∧ 𝑎∗ 𝜑(𝐿) = [𝑎 ∨ 𝑎∗ ) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎 ∈ 𝐵(𝐿).
-6-
Corollary 1:
Let 𝐿 be a quasi-modular 𝑆-algebra. Then the structure
map 𝜑(𝐿): 𝐵(𝐿) ⟶ 𝐹(𝐷(𝐿)) is a {0,1}-homomorphism of 𝐵(𝐿) into
(𝐷(𝐿)) .
We can assign to each quasi modular 𝑆-algebra 𝐿 a triple <
𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿) > where
𝑎𝜑(𝐿) = [𝑎∗ ) ∩ 𝐷(𝐿), 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎 ∈ 𝐵(𝐿).
For more information of the triple of a quasi modular palgebra we refer to [8].
Lemma 2:
If 𝐿 is a quasi-modular double 𝑆-algebra, then for any 𝑥 ∈ 𝐿
𝑥 + = 𝑥 ∗ ∨ (𝑥 ∨ 𝑥 ∗ )++∗
Proof:
Since for any 𝑥 ∈ 𝐿
𝑥 = 𝑥 ∗∗ ∧ (𝑥 ∨ 𝑥 ∗ )
So,
+
𝑥 + = (𝑥 ∗∗ ∧ (𝑥 ∨ 𝑥 ∗ )) = 𝑥 ∗∗+ ∨ (𝑥 ∨ 𝑥 ∗ )+ = 𝑥 ∗ ∨ (𝑥 ∨ 𝑥 ∗ )+
= 𝑥 ∗ ∨ (𝑥 ∨ 𝑥 ∗ )++∗
Lemma 3:
If 𝐿, 𝐿1 are quasi-modular double 𝑆-algebras, and ℎ: 𝐿 ⟶ 𝐿1
be a {*} - homomorphism. Then ℎ is a {+} - homomorphism if and
only if 𝑑 ++ ℎ = (𝑑ℎ)++ , for all 𝑑 ∈ 𝐷(𝐿).
-7-
Proof:
Let ℎ be a {+}-homomorphism, then 𝑥 + ℎ = (𝑥ℎ)+
For any 𝑥 ∈ 𝐿. So,
𝑑 ++ ℎ = (𝑑 + ℎ)+ = (𝑑ℎ)++
Conversely, for any 𝑥 ∈ 𝐿, we have
𝑥 = 𝑥 ∗∗ ∧ (𝑥 ∨ 𝑥 ∗ ) 𝑎𝑛𝑑 𝑥 + = 𝑥 ∗ ∨ (𝑥 ∨ 𝑥 ∗ )++∗ ,
Then
𝑥 + ℎ = (𝑥 ∗ ∨ (𝑥 ∨ 𝑥 ∗ )++∗ )ℎ = (𝑥 ∗ ∨ 𝑑 + )ℎ,
𝑑 = 𝑥 ∨ 𝑥∗
= (𝑥 ∗ ∨ 𝑑 ++∗ )ℎ = 𝑥 ∗ ℎ ∨ 𝑑 ++∗ ℎ = (𝑥ℎ)∗ ∨ (𝑑++ ℎ)∗
For ℎ is a {*}-homomorphism and 𝑑 ++ ℎ = (𝑑ℎ)++ is given. Thus,
𝑥 + ℎ = (𝑥ℎ)∗ ∨ (𝑑 ++ ℎ)∗ = (𝑥ℎ)∗ ∨ (𝑑ℎ)++∗
= (𝑥ℎ)∗ ∨ (𝑑ℎ)+ = (𝑥ℎ)∗∗+ ∨ (𝑑ℎ)+
= ((𝑥ℎ)∗∗ ∧ 𝑑ℎ)+ = (𝑥ℎ)+ .
Theorem 2:
If 𝐿 is a quasi-modular 𝑆-algebra. Then 𝐿 is a quasi-modular
double 𝑆-algebra if and only if there exists a {1}-homomorphism
𝜓(𝐿) ∶ 𝐷(𝐿) ⟶ 𝐵(𝐿) such that
(𝑖)𝑑 ≥ 𝑑𝜓(𝐿),
(𝑖𝑖)𝑑 ≥ 𝑎 , 𝑎 ∈ 𝐵(𝐿) 𝑎𝑛𝑑 𝑑 ∈ 𝐷(𝐿) 𝑖𝑚𝑝𝑙𝑦 𝑡ℎ𝑎𝑡 𝑑𝜓(𝐿) ≥ 𝑎 .
-8-
Proof:
Let 𝐿 is a quasi-modular 𝑆-algebra. Define
𝜓(𝐿) ∶ 𝐷(𝐿) ⟶ 𝐵(𝐿) by 𝑑𝜓(𝐿) = 𝑑 ++
𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑑 ∈ 𝐷(𝐿)
The map 𝜓(𝐿) is a {1}-homomorphism because 1𝜓(𝐿) = 1++ , and
𝑑 ≥ 𝑑 ++ = 𝑑𝜓(𝐿) .
Also,
𝑑 ≥ 𝑎 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑑++ ≥ 𝑎++ 𝑖. 𝑒 𝑑𝜓(𝐿) ≥ 𝑎, 𝑎 ∈ 𝐵(𝐿).
Hence, 𝜓(𝐿) satisfies(𝑖)𝑎𝑛𝑑 (𝑖𝑖) .
Conversely, suppose that there is a {1}-homomorphism
𝜓(𝐿) ∶ 𝐷(𝐿) ⟶ 𝐵(𝐿) satisfies (𝑖)𝑎𝑛𝑑 (𝑖𝑖). Let 𝑡 ∈ 𝐿, we shall prove
that
𝑡 + = 𝑡 ∗ ∨ ((𝑡 ∨ 𝑡 ∗ )𝜓(𝐿))∗ is a dual pseudocomplemented
element of 𝑡 ∈ 𝐿. Put, 𝑎 = 𝑡 ∗∗ 𝑎𝑛𝑑 𝑑 = 𝑡 ∨ 𝑡 ∗ , then
∗
𝑡 ∨ 𝑡 + = (𝑎 ∧ 𝑑) ∨ (𝑎∗ ∨ (𝑑𝜓(𝐿))
≥ (𝑎 ∧ 𝑑𝜓(𝐿)) ∨ (𝑎∗ ∨ (𝑑𝜓(𝐿))
∗
∗
∗
≥ (𝑎 ∨ 𝑎∗ ∨ (𝑑𝜓(𝐿)) ) ∧ (𝑑𝜓(𝐿) ∨ 𝑎∗ ∨ (𝑑𝜓(𝐿)) )
≥ 1 ∧ 1 = 1. (holds because all elements are in 𝐵(𝐿)).
So, 𝑡 ∨ 𝑡 + = 1.
Finally, for , 𝑣 ∈ 𝐿, 𝑡 = 𝑎 ∧ 𝑑, 𝑣 = 𝑏 ∧ 𝑒 , where
𝑎 = 𝑡 ∗∗ ,
𝑏 = 𝑣 ∗∗ ,
𝑑 = 𝑡 ∨ 𝑡 ∗ 𝑎𝑛𝑑 𝑒 = 𝑣 ∨ 𝑣 +
We shall prove that ∨ 𝑣 = 1 𝑖𝑓𝑓 𝑣 ≥ 𝑡 + .
Since 𝑡 ∨ 𝑣 = 1 , then
-9-
1 = (𝑎 ∧ 𝑑) ∨ (𝑏 ∧ 𝑒) ≤ (𝑎 ∧ 𝑑 𝜓(𝐿)) ∨ (𝑏 ∧ 𝑒𝜓(𝐿))
≤ ((𝑎 ∧ 𝑑 𝜓(𝐿)) ∨ 𝑏) ∧ (𝑎 ∧ 𝑑𝜓(𝐿) ∨ 𝑒𝜓(𝐿))
(holds because all elements are in 𝐵(𝐿))
≤ (𝑎 ∨ 𝑏)(𝑏 ∨ 𝑑𝜓(𝐿)) ∧ (𝑎 ∨ 𝑒𝜓(𝐿)) ∧ (𝑑𝜓(𝐿) ∨ 𝑒𝜓(𝐿))
This implies that
1 = (𝑎 ∨ 𝑏) ∧ (𝑏 ∨ 𝑑𝜓(𝐿)) ∧ (𝑎 ∨ 𝑒𝜓(𝐿)) ∧ (𝑑𝜓(𝐿) ∨ 𝑒𝜓(𝐿))
So 1 = (𝑎 ∨ 𝑏) , 1 = (𝑏 ∨ 𝑑𝜓(𝐿)) , 1 = (𝑎 ∨ 𝑒 𝜓(𝐿)) and
1 = 𝑑𝜓(𝐿) ∨ 𝑒𝜓(𝐿)
Since, 𝑎, 𝑏, 𝑑𝜓(𝐿), 𝑒𝜓(𝐿) ∈ 𝐵(𝐿), then
1 = (𝑎 ∨ 𝑏)
𝑖𝑚𝑝𝑙𝑖𝑒𝑠
𝑏 ≥ 𝑎∗ , and
1 = (𝑏 ∨ 𝑑𝜓(𝐿)) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑑𝜓(𝐿) ≥ 𝑏 ∗ , 𝑖. 𝑒 𝑑 ≥ 𝑏 ∗ , and
1 = (𝑎 ∨ 𝑒𝜓(𝐿)) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠
𝑒𝜓(𝐿) ≥ 𝑎∗ 𝑖. 𝑒 𝑒 ≥ 𝑎∗ , and
∗
1 = 𝑑𝜓(𝐿) ∨ 𝑒𝜓(𝐿) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑒𝜓(𝐿) ≥ (𝑑𝜓(𝐿)) 𝑖. 𝑒 𝑒 ≥ (𝑑𝜓(𝐿))∗ .
By using Lemma 1, we get
𝑡 + = 𝑡 ∗ ∨ (𝑡 ∨ 𝑡 ∗ )++∗ = 𝑎∗ ∨ (𝑑𝜓(𝐿))∗ ≤ 𝑏 ∨ 𝑒 = 𝑣
Hence, 𝐿 is a double p-algebra. The form of 𝑡 ++ is given by
𝑡 + = 𝑎∗ ∨ (𝑑𝜓(𝐿))∗ 𝑖𝑚𝑝𝑙𝑖𝑒𝑠
𝑡 ++ = (𝑎∗ ∨ (𝑑𝜓(𝐿))∗ )+ and
𝑡 ++ = 𝑎∗+ ∧ (𝑑𝜓(𝐿))∗+ = 𝑎∗∗ ∧ (𝑑𝜓(𝐿))∗∗ = 𝑎 ∧ 𝑑𝜓(𝐿)
So,
𝑡 + ∧ 𝑡 ++ = (𝑎∗ ∨ (𝑑𝜓(𝐿))∗ ) ∧ ( 𝑎 ∧ 𝑑𝜓(𝐿))
- 10 -
𝑡 + ∧ 𝑡 ++ = (𝑎∗ ∧ ( 𝑎 ∧ 𝑑𝜓(𝐿)) ∨ ((𝑑𝜓(𝐿))∗ ) ∧ ( 𝑎 ∧ 𝑑𝜓(𝐿)))
𝑡 + ∧ 𝑡 ++ = (𝑎∗ ∧ 𝑎 ∧ 𝑑𝜓(𝐿)) ∨ ((𝑑𝜓(𝐿))∗ ) ∧ 𝑑𝜓(𝐿) ∧ 𝑎))
𝑡 + ∧ 𝑡 ++ = 0 ∨ 0
It remains to prove the dual quasi-modularity of 𝐿, we want to
prove
(𝑥 ∧ 𝑡 ++ ) ∨ 𝑥 = 𝑦 ∧ (𝑥 ∨ 𝑡 ++ ) , 𝑥 ≤ 𝑦
(2)
We have 𝑦 ∧ (𝑥 ∨ 𝑡 ++ ) = (𝑦 ∧ (𝑎 ∧ 𝑑𝜓(𝐿)) ∨ 𝑥
∗∗
= (𝑦 ∧ (𝑎 ∧ 𝑑𝜓(𝐿)) ∨ 𝑥
∗∗
= (𝑥 ∨ (𝑎 ∧ 𝑑𝜓(𝐿)) ∧ 𝑦 = (𝑥 ∨ 𝑡 ++ ) ∧ 𝑦
Because 𝑎, 𝑑𝜓(𝐿) ∈ 𝐵(𝐿). Hence, 𝐿 is a quasi-modular double 𝑆algebra.
Combine Theorem 1 and Theorem 2, we get the quadruples
< 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >
which
is
called
the
quadruple
associated with a quasi-modular double 𝑆-algebra 𝐿, and
determines all of 𝐿.
In a quasi-modular double p-algebra. Let
𝐷𝑎 = {𝑥 ∈ 𝑎𝜑(𝐿) ∶ 𝑥 ≤ 𝑎 ∨ 𝑎∗ }, 𝑎 ∈ 𝐵(𝐿)
we can prove that the isomorphism between 𝐹𝑎 and 𝐷𝑎 given by
𝑥 ⟶ 𝑥 ∨ 𝑎∗
Hence, we can identify an element 𝑥 ∈ 𝐿, 𝑥 ∈ 𝐹𝑎 with the
ordered pair
- 11 -
(𝑎, 𝑎∗ 𝜑(𝐿) ∨ [𝑥 ∨ 𝑎∗ ) ∈ 𝐵(𝐿) × 𝐹(𝐷(𝐿)).
Thus, 𝑎∗ 𝜑(𝐿) ∨ [𝑥 ∨ 𝑎∗ ) = ([𝑎) ∩ 𝐷(𝐿))𝜑(𝐿) ∨ ([𝑥) ∧ [𝑎∗ ))
= 𝑎∗ 𝜑(𝐿) ∨ [𝑥)
Also, if 𝑥 = (𝑎, 𝑎∗ 𝜑(𝐿) ∨ [𝑥 ∨ 𝑎∗ )), 𝑦 = (𝑏, 𝑏 ∗ 𝜑(𝐿) ∨ [𝑦 ∨ 𝑏 ∗ )), then
𝑥 ≤ 𝑦 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓
𝑎 ≤ 𝑏 and 𝑎∗ 𝜑(𝐿) ∨ [𝑥 ∨ 𝑎∗ ) ⊇ 𝑏 ∗ 𝜑(𝐿) ∨ [𝑦 ∨ 𝑏 ∗ )
So, we prove
Theorem 3 (Uniqueness)
A quasi-modular double 𝑆-algebra 𝐿 is determined up to
isomorphism by the quadruple < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >.
From the above discussion we assert that the quadruple <
𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) > associated with a qusi-modular double 𝑆algebra 𝐿 is a quasi-modular quadruple.
4: Construction of quasi-modular double S-algebra:
Definition 1:[8]
An 𝐹- triple < 𝐵, 𝐷, 𝜑 > is said to be a quasi-modular 𝐹-triple
if
(𝑖) < 𝐵;∧,∨, ′, 0,1 > is a Boolean algebra,
(𝑖𝑖) < 𝐷;∩,∪ ,1 > is a lattice with
,
(𝑖𝑖𝑖) 𝜑 is a {0,1,∨}-homomorphism from 𝐵 into 𝐹(𝐷) such that for
𝑎 ∈ 𝐵 there is an element 𝑑𝑎 ∈ 𝐷 with 𝑎𝜑 ∩ 𝑎′𝜑 = [𝑑𝑎 ),
- 12 -
(𝑣) 𝑖𝑓 𝑎, 𝑏, 𝑐, 𝑔 ∈ 𝐵 𝑎𝑛𝑑 𝑥, 𝑦, 𝑧 ∈ 𝐷 with 𝑎 ∧ 𝑏 ≥ 𝑐, and
(𝑎𝜑 ∪ [𝑥)) ∩ (𝑏𝜑 ∪ [𝑦)) ⊇ (𝑐𝜑 ∪ [𝑧)). Then
(𝑎𝜑 ∪ [𝑥)) ∩ (𝑏𝜑 ∪ [𝑦)) ∩ (𝑔𝜑 ∪ 𝑐𝜑 ∪ [𝑧))
= ((𝑎𝜑 ∪ [𝑥)) ∩ (𝑏𝜑 ∪ [𝑦)) ∩ (𝑔𝜑 ∪ (𝑐𝜑 ∪ [𝑧))
Definition 2:
A quasi-modular 𝐹-triple < 𝐵, 𝐷, 𝜑 > is said to be a quasimodular 𝑆-triple if 𝜑 is a {0,1}-homomorphism map from 𝐵 into
𝐹(𝐷).
Definition 3: [8]
An isomorphism of the quasi-modular 𝑆-triples < 𝐵, 𝐷, 𝜑 >
and < 𝐵1 , 𝐷1 , 𝜑1 > is a pair (𝑓, 𝑔) where 𝑓 is an isomorphism
from 𝐵 into 𝐵1 and 𝑔 is an isomorphism from 𝐷 into 𝐷1 such the
diagram
𝐵
𝑓↓
𝐵1
𝜑
→
𝐹(𝐷)
↓ 𝐹(𝑔)
𝜑1
→
𝐹(𝐷1 )
is commutative. i.e. 𝜑𝐹(𝑔) = 𝑓𝜑1 , where 𝐹(𝑔) stands for the
isomorphism of 𝐹(𝐷) 𝑎𝑛𝑑 𝐹(𝐷1 ) induced by 𝑔.
Definition 4:
A quadruple < 𝐵, 𝐷, 𝜑, 𝜓 > is said to be a quasi-modular
quadruple if the 𝑆-triple < 𝐵, 𝐷, 𝜑 > is a quasi-modular 𝑆-triple and
𝜓 is a {1}-homomorphism of 𝐷 into 𝐵 such that
(1) 𝑑 ∈ (𝑑𝜓)′ 𝜑 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑑 ∈ 𝐷,
- 13 -
(2) 𝑑 ∈ 𝑎′ 𝜑, 𝑎 ∈ 𝐵 𝑖𝑚𝑝𝑙𝑦 𝑡ℎ𝑎𝑡 𝑑𝜓 ≥ 𝑎.
Theorem 4 : (Construction Theorem)
Let < 𝐵, 𝐷, 𝜑, 𝜓 > be a quasi-modular quadruple. Then we
construct a quasi-modular double 𝑆-algebra 𝐿 such that the
quadruple𝑠 < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝑆) >
and
< 𝐵, 𝐷, 𝜑, 𝜓 >
are
isomorphic.
Proof:
By the definition of the quasi-modular quadruple, we have <
𝐵, 𝐷, 𝜑 >
is a quasi-modular 𝑆-triple and satisfies the conditions
(1),(2) as in Theorem 2. By the Construction Theorem of quasimodular 𝑆-algebra [8, Theorem 4,6.5], we get a quasi-modular 𝑆algebra 𝐿 such that the 𝑆-triples
< 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿) > 𝑎𝑛𝑑
< 𝐵, 𝐷, 𝜑 > are isomorphic. So, it is enough to prove the
conditions (1),(2) as in Theorem 2, where
𝐿 = {(𝑎, 𝑎′ 𝜑 ∨ [𝑑)): 𝑎 ∈ 𝐵, 𝑑 ∈ 𝐷},
𝐵(𝐿) = {(𝑎, 𝑎′ 𝜑): 𝑎 ∈ 𝐵},
𝐷(𝐿) = {(1, [𝑑)): 𝑑 ∈ 𝐷}.
Define 𝜓(𝐿) = 𝑔𝜓𝑓 −1 : 𝐷(𝐿) ⟶ 𝐵(𝐿) 𝑏𝑦
𝑑1 𝑔𝜓𝑓 −1 = < 𝑑𝜓, (𝑑𝜓)′ 𝜑 > , 𝑑1 =< 1, [𝑑) > 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑑1 ∈ 𝐷(𝐿).
Since, 𝑑 ∈ (𝑑𝜓)′ 𝜑 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑑 ∈ 𝐷 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 [𝑑) ⊆ (𝑑𝜓)′ 𝜑
So, < 1, [𝑑) > ≥ < 𝑑𝜓, (𝑑𝜓)′ 𝜑 >, that is 𝑑1 ≥ 𝑑1 𝑔𝜓𝑓 −1 =
𝑑1 𝜓(𝐿).
- 14 -
Also, 𝑑 ∈ 𝑎′ 𝜑, 𝑎 ∈ 𝐵 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 𝑑𝜓 ≥ 𝑎 and so, (𝑑𝜓)′ ≤ 𝑎′ then
we get (𝑑𝜓)′ 𝜑 ≤ 𝑎′ 𝜑. Thus < 𝑑𝜓, (𝑑𝜓)′ 𝜑 > ≥ < 𝑎, 𝑎′ 𝜑 > that is
𝑑1 𝜓(𝐿) ≥ 𝑎1 , and 𝜓(𝐿) = 𝑔𝜓𝑓 −1 . Hence,
< 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >
Is a quadruple associated with the quasi-modular double 𝑆-algebra
𝐿, and < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) > isomorphic with < 𝐵, 𝐷, 𝜑, 𝜓 > ?
Corollary 1:
A double 𝑆-algebra 𝐿 is a quasi-modular double 𝑆-algebra if
and only if the quadruple < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >.
Proof:
The proof follows from Theorems 2,3 in the last section and
Theorem 4.
Definition 5:
An isomorphism of the quasi-modular quadruple
< 𝐵, 𝐷, 𝜑, 𝜓 > and < 𝐵1 , 𝐷1 , 𝜑1 , 𝜓1 > is a pair < 𝑓, 𝑔 > where 𝑓 is
an isomorphism of 𝐵 and 𝐵1 and 𝑔 is an isomorphism of 𝐷 and 𝐷1
such that the diagrams
𝐵
𝑓↓
𝐵1
𝜑
→
𝜑(𝐿)
→
𝐹(𝐷)
↓ 𝐹(𝑔)
𝐹(𝐷1 )
𝐷
𝑔↓
𝐷1
are commutative.
- 15 -
𝜓
→
𝜓1
→
𝐵
↓𝑓
𝐵1
Theorem 5:
Two quasi-modular quadruples < 𝐵, 𝐷, 𝜑, 𝜓 > and
< 𝐵1 , 𝐷1 , 𝜑1 , 𝜓1 > are isomorphic if and only if the quasi-modular 𝑆triples < 𝐵, 𝐷, 𝜑 > and < 𝐵1 , 𝐷1 , 𝜑1 >are isomorphic.
Proof:
The necessary condition is obvious.
Conversely, suppose that the quasi-modular 𝑆- triples <
𝐵, 𝐷, 𝜑 > and < 𝐵1 , 𝐷1 , 𝜑1 > are isomorphic, then by the
Construction Theorem of quasi-modular 𝑆-triple [8,Theorem 4,6.5)
there exists a qusi-modular 𝑆-algebra 𝐿 such that the triples <
𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿) > and < 𝐵, 𝐷, 𝜑 > are isomorphic, and also there
exists a qusi-modular 𝑆-algebra 𝐿1 such that the triples <
𝐵1 (𝐿), 𝐷1 (𝐿), 𝜑1 (𝐿) > and < 𝐵1 , 𝐷1 , 𝜑1 > are isomorphic.
Let < 𝑓, 𝑔 > be the isomorphism between the triples <
𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿) > and < 𝐵, 𝐷, 𝜑 > . Define the map
𝑔𝜓𝑓 −1 : 𝐷(𝐿) ⟶ 𝐵(𝐿) 𝑏𝑦
𝑑𝑔𝜓𝑓 −1 = 𝑑 ++ 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑑 ∈ 𝐷(𝐿).
We show that 𝑔𝜓𝑓 −1 satisfies the conditions (1),(2) of
Theorem 2, as the following
1𝑔𝜓 = 1𝜓 = 1++ = 1 = 1𝑓 𝑖𝑚𝑝𝑙𝑖𝑒𝑠
1 𝑔𝜓𝑓 −1 = 1
Since 𝑓, 𝑔 are isomorphic, let 𝑑1 = 𝑑𝑔
𝑎𝑛𝑑
𝑎1 =
𝑎𝑓, 𝑤ℎ𝑒𝑟𝑒 𝑑1 ∈ 𝐷1 , 𝑎1 ∈ 𝐵1 , 𝑑 ∈ 𝐷, 𝑎 ∈ 𝐵. Then we have
𝑑𝑔𝜓1 𝑓 −1 = 𝑑 ++
- 16 -
So, 𝑑𝑔𝜓1 𝑓 −1 = 𝑑 ++ . So, 𝑑 ≥ 𝑑 ++ = 𝑑𝑔𝜓1 𝑓 −1 .
Moreover, if 𝑑 ∈ 𝐷, 𝑎 ∈ 𝐵 𝑠𝑢𝑐ℎ 𝑡ℎ𝑎𝑡 𝑑 ≥ 𝑎, we have
𝑑1 ≥ 𝑎1 that is 𝑑1++ ≥ 𝑎1++ , thus 𝑑𝑔𝜓1 ≥ 𝑎1++ = 𝑎1 = 𝑎𝑓, therefore
𝑑𝑔𝜓1 𝑓 −1 ≥ 𝑎++ .
Hence, the corresponding quadruples are isomorphic.
< 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) > is a quadruple associated with the
quasi-modular double 𝑆-algebra 𝐿, and < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >
isomorphic with < 𝐵, 𝐷, 𝜑, 𝜓 > .
Theorem 6:
Two quasi-modular 𝑆-algebras are isomorphic if and only if
the associated quadruples are isomorphic. Every quasi-modular
quadruple is isomorphic to a quadruple associated with double 𝑆algebra.
Proof:
The proof follows from the uniqueness of quasi-modular
double 𝑆-algebra Theorem 1 and Corollary 1.
5: Homomorphisms and Subalgebras:
Let 𝐿 and 𝐿1 be a quasi-modular double 𝑆-algebra and a
map ℎ: 𝐿 → 𝐿1 be a homomorphism of 𝐿 into 𝐿1 i.e
preserves
the operations ∧,∨ ,*, +,0,1.
If ℎ: 𝐿 → 𝐿1 is a homomorphism. Then the restriction ℎ𝐵 =
ℎ|𝐵(𝐿) is a Boolean homomorphism of 𝐵(𝐿) into 𝐵(𝐿1 ) and the
- 17 -
restriction ℎ𝐷 = ℎ|𝐷(𝐿) is a homomorphism of 𝐷(𝐿) into 𝐷(𝐿1 ) that
preserves
Theorem 7:
Let 𝐿 𝑎𝑛𝑑 𝐿1 be quasi-modular 𝑆-algebras and
be a
homomorphism of 𝐿 onto 𝐿1 . Then, the restriction ℎ𝐵 = ℎ|𝐵(𝐿) is a
homomorphism of 𝐵(𝐿) onto 𝐵(𝐿1 ) and the restriction ℎ𝐷 = ℎ|𝐷(𝐿)
is a homomorphism of 𝐷(𝐿) onto 𝐷(𝐿1 ) that preserves .
Proof:
The proof is straightforward from [8,7.1] and the
Characterization Theorem .
Definition 7: [8]
A homomorphism of the quasi-modular 𝑆-triples < 𝐵, 𝐷, 𝜑 >
into a quasi-modular 𝑆-triple < 𝐵1 , 𝐷1 , 𝜑1 > is a pair (𝑓, 𝑔), where 𝑓
is a Boolean homomorphism of 𝐵 into 𝐵1 , 𝑔 is a homomorphism of
𝐷 into 𝐷1 .preserving
such that for every 𝑎 ∈ 𝐵
𝑑𝑎 𝑔 = 𝑑𝑎𝑓 ,
𝑑𝑎 = 𝑎 ∨ 𝑎′
Theorem 8: [8, Theorem 5]
Let
𝑆
ℎ: 𝐿 → 𝐿1
-algebras.
Then,
be
homomorphism
the pair (ℎ𝐵 , ℎ𝐷 )
of
quasi-modular
is a
quasi-modular
𝑆 -triple homomorphism between the corresponding associated
triples.Conversely,every quasi-modular 𝑆-triple homomorphism
(𝑓, 𝑔): < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿) > → < 𝐵(𝐿1 ), 𝐷(𝐿1 ), 𝜑(𝐿1 ) >
- 18 -
uniquely determines a homomorphism
ℎ: 𝐿 → 𝐿1 such that
ℎ𝐵 = 𝑓 𝑎𝑛𝑑 ℎ𝐷 = 𝑔 according to the following rule
𝑥ℎ = 𝑥 ∗∗ 𝑓 ∧ 𝑑𝑔 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐿.
Definition 7:
Let < 𝐵, 𝐷, 𝜑, 𝜓 > and < 𝐵1 , 𝐷1 , 𝜑1 , 𝜓1 > be a quasi-modular
quadruples. A homomomrphism of the quadruple < 𝐵, 𝐷, 𝜑, 𝜓 >
into a quadruple< 𝐵1 , 𝐷1 , 𝜑1 , 𝜓1 > is a pair (𝑓, 𝑔), where 𝑓 is a
Boolean homomorphism of 𝐵 into 𝐵1 , 𝑔 is a homomorphism of 𝐷
into 𝐷1 .preserving
such that for every 𝑎 ∈ 𝐵
𝑎𝜑𝑔 ⊆ 𝑎𝑓𝜑1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑎 ∈ 𝐵 and
𝑑𝜓𝑓 = 𝑑𝑔𝜓1 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑑 ∈ 𝐷
Theorem 9:
Let 𝐿 and 𝐿1 be quasi-modular 𝑆-algebras, < 𝐵, 𝐷, 𝜑, 𝜓 >
and < 𝐵1 , 𝐷1 , 𝜑1 , 𝜓1 >be the associated quadruples, respectively.
Let ℎ be a homomorphism of 𝐿 and 𝐿1 . Then (ℎ𝐵 , ℎ𝐷 ) is a
homomorphism of the quadruples. Conversely, every
homomorphism (𝑓, 𝑔) of the quadruples uniquely determines a
homomorphism ℎ: 𝐿 → 𝐿1 with ℎ𝐵 = 𝑓 and ℎ𝐷 = 𝑔 according to the
following rule
𝑥ℎ = 𝑥 ∗∗ 𝑓 ∧ (𝑥 ∨ 𝑥 ∗ )𝑔 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝐿
Proof:
The proof follows from [5], [8] and Lemma1.
- 19 -
Lemma 4:
Let 𝐿1 be a subalgebra of the quasi-modular double 𝑆algebra 𝐿. Then 𝐵1 = 𝑆1 ∩ 𝐵(𝑆) is a subalgebra of 𝐵(𝑆) and 𝐷1 =
𝐷1 ∩ 𝐷(𝑆) is a sublattice of 𝐷(𝑆) containing
associated with 𝐿1 is
𝑓𝑜𝑟 𝑎 ∈ 𝐵1 and
. The quadruple
< 𝐵1 , 𝐷1 , 𝜑1 , 𝜓1 > where 𝑎𝜑1 = 𝑎𝜑 ∩ 𝐷1 ,
𝑑𝜓1 = 𝑑𝜓 ∩ 𝐵1 , 𝑓𝑜𝑟 𝑎 ∈ 𝐷1 .
Proof:
The proof is straightforward.
Theorem 10:
Let 𝐿 be a quasi-modular double 𝑆-algebra, 𝐵1 is a
subalgebra of 𝐵(𝐿), 𝐷1 is a subalgebra of 𝐷(𝐿) containing 1. We
can fill in < 𝐵1 , 𝐷1 , ? , ? > such that it will become the quadruple
associated with a subalgebra of 𝐿 if and only if
(𝑎𝜑(𝐿) ∩ 𝐷1 ) ∪ (𝑎∗ 𝜑(𝐿) ∩ 𝐷1 ) = 𝐷1 𝑓𝑜𝑟 𝑎 ∈ 𝐵1 ,
𝑑 ++ ∈ 𝐵1 𝑓𝑜𝑟 𝑒𝑣𝑒𝑟𝑦 𝑑 ∈ 𝐷1
Proof:
The proof follows from the Construction Theorem and [8,6.5
vi] and the above Lemma, we have 𝜑1 = 𝜑 ∩ 𝐷1 and 𝜓1 = 𝜓 ∩ 𝐵1
such that the quadruple < 𝐵1 , 𝐷1 , 𝜑1 , 𝜓1 > associated with a
subalgebra of 𝐿.
- 20 -
6 : Congruence Relation
Definition 7:
Let
< 𝐵, 𝐷, 𝜑, 𝜓 >
be
a
qusi-modular
quadruple.
A
congruence relation of < 𝐵, 𝐷, 𝜑, 𝜓 > is a pair (𝜃𝐵 , 𝜃𝐷 ) where 𝜃𝐵 is
a congruence relation of 𝐵, 𝜃𝐷 is a congruence relation of 𝐷 and
(1) 𝑎 ≡ 0(𝜃𝐵 ) implies 𝑥 ≡ 1(𝜃𝐷 ) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑥 ∈ 𝑎𝜑,
(2) 𝑑 ≡ 𝑒(𝜃𝐷 ) implies 𝑑𝜓 ≡ 𝑒𝜓(𝜃𝐵 ) 𝑓𝑜𝑟 𝑎𝑙𝑙 𝑑 ∈ 𝐷
hold.
Let 𝜃 be a congruence relation of a qusi-modular double 𝑆algebra 𝐿. Denote 𝜃𝐵(𝐿) and 𝜃𝐷(𝐿) , the restirction of 𝜃 to 𝐵(𝐿) ×
𝐵(𝐿) and 𝐷(𝐿) × 𝐷(𝐿), respectively.
Theorem 11:
Let 𝐿 be a quasi-modular double 𝑆-algebra. If 𝜃 is a
congruence relation of 𝐿 then (𝜃𝐵(𝐿) , 𝜃𝐷(𝐿) ) is a congruence of the
quadruple < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >. Conversely, let (𝜃𝐵(𝐿) , 𝜃𝐷(𝐿) )
be a congruence of the quadruple < 𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >.
Then there exists a uniquely determined a congruence 𝜃 of 𝐿 with
𝜃𝐵(𝐿) = 𝜃𝐵 and 𝜃𝐷(𝐿) = 𝜃𝐷 such that
𝑥 ≡ 𝑦(𝜃) 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 ∗∗ ≡ 𝑦 ∗∗ (𝜃𝐵 ) 𝑎𝑛𝑑 𝑥 ∨ 𝑥 ∗ ≡ 𝑦 ∨ 𝑦 ∗ (𝜃𝐷 ).
Proof:
The first statement of theorem is evident. Conversely,
suppose we have a congruence (𝜃𝐵(𝐿) , 𝜃𝐷(𝐿) ) of the quadruple <
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𝐵(𝐿), 𝐷(𝐿), 𝜑(𝐿), 𝜓(𝐿) >. By [83, Theorem6], there exists a
uniquely congruence 𝜃 of 𝐿 such that 𝜃𝐵(𝐿) = 𝜃𝐵 , 𝜃𝐷(𝐿) = 𝜃𝐷 and
𝑥 ≡ 𝑦(𝜃) 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 ∗∗ ≡ 𝑦 ∗∗ (𝜃𝐵 ) 𝑎𝑛𝑑 𝑥 ∨ 𝑥 ∗ ≡ 𝑦 ∨ 𝑦 ∗ (𝜃𝐷 ).
We prove 𝜃 is a + congruence, that is
𝑥 ≡ 𝑦(𝜃) 𝑖𝑓 𝑎𝑛𝑑 𝑜𝑛𝑙𝑦 𝑖𝑓 𝑥 + ≡ 𝑦 + (𝜃).
Since, 𝑥 + = 𝑥 ∗ ∨ (𝑥 ∨ 𝑥 ∗ )++∗ and 𝑦 + = 𝑦 ∗ ∨ (𝑦 ∨ 𝑦 ∗ )++∗ . By
Lemma 1, we have 𝑥 ∗ ≡ 𝑦 ∗ (𝜃) 𝑖𝑚𝑝𝑙𝑖𝑒𝑠 𝑡ℎ𝑎𝑡 (𝑥 ∨ 𝑥 ∗ )++∗ ≡
(𝑦 ∨ 𝑦 ∗ )++∗ (𝜃). By hypothesis, we obtain 𝑥 + ≡ 𝑦 + (𝜃).
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