Dual skew Heyting almost distributive lattices

[9] Let H be an ADL. Then for any w, x, y, z ∈ H, we have the following.

(1)∧isassociative(2)x∧y∧z=y∧x∧z(3)(x∨y)∧z=(y∨x)∧z(4){ w∨(x∨y) }∧z={ (w∨x)∨y }∧z(5)Ifx≤zandy≤z,thenx∧y=y∧xandx∨y=y∨x. $$\begin{align}& \left( 1 \right)\wedge is\,associative \\ & \left( 2 \right)x\wedge y\wedge z=y\wedge x\wedge z \\ & \left( 3 \right)\left( x\vee y \right)\wedge z=\left( y\vee x \right)\wedge z \\ & \left( 4 \right)\left\{ w\vee \left( x\vee y \right) \right\}\wedge z=\left\{ \left( w\vee x \right)\vee y \right\}\wedge z \\ & \left( 5 \right)If\,x\le z\,and\,y\le z,\,then\,x\wedge y=y\wedge x\,and\,\,x\vee y=y\vee x. \\ \end{align}$$

Definition 2.3. [3] An algebra (H,∨,∧,→,0,1) of type (2,2,2,0,0) is called a Heyting algebra if it satisfies the following conditions.

(1)(H,∨,∧,0,1)isaboundeddistributivelattice(2)x→x=1(3)y≤x→y(4)x∧(x→y)=x∧y(5)x→(y∧z)=(x→y)∧(x→z)(6)(x∨y)→z=(x→z)∧(y→z) $$\begin{align}& \left( 1 \right)\,\left( H,\vee ,\wedge ,0,1 \right)\,\text{is}\,\text{a}\,\text{bounded}\,\text{distributive}\,\text{lattice} \\ & \left( 2 \right)x\to x=1 \\ & \left( 3 \right)y\le x\to y \\ & \left( 4 \right)x\wedge \left( x\to y \right)=x\wedge y \\ & \left( 5 \right)x\to \left( y\wedge z \right)=\left( x\to y \right)\wedge \left( x\to z \right) \\ & \left( 6 \right)\left( x\vee y \right)\to z=\left( x\to z \right)\wedge \left( y\to z \right) \\ \end{align}$$

for all x, y, z ∈ H.

[3] Let H be a Heyting algebra, then an equivalence relation θ on H is a congruence relation if and only if for any (a,b) ∈ θ,d ∈ H,

(1)(a∧d,b∧d)∈θ(2)(a∨d,b∨d)∈θ(3)(a→d,b→d)∈θ(4)(d→a,d→b)∈θ. $$\begin{align}& \left( 1 \right)\left( a\wedge d,\,b\wedge d \right)\in \theta \\ & \left( 2 \right)\left( a\vee d,\,b\vee d \right)\in \theta \\ & \left( 3 \right)\left( a\to d,\,b\to d \right)\in \theta \\ & \left( 4 \right)\left( d\to a,d\to b \right)\in \theta . \\ \end{align}$$

Definition 2.5. [3] Let (H,∨,∧,0,m) be an ADL with 0 and a maximal element m. Suppose → is a binary operation on H satisfying the following conditions.

(1)x→x=m(2)(x→y)∧y=y(3)x∧(x→y)=x∧y∧m(4)x→(y∧z)=(x→y)∧(x→z)(5)(x∨y)→z=(x→z)∧(y→z) $$\begin{align}& \left( 1 \right)\,x\to x=m \\ & \left( 2 \right)\left( x\to y \right)\wedge y=y \\ & \left( 3 \right)x\wedge \left( x\to y \right)=x\wedge y\wedge m \\ & \left( 4 \right)x\to \left( y\wedge z \right)=\left( x\to y \right)\wedge \left( x\to z \right) \\ & \left( 5 \right)\left( x\vee y \right)\to z=\left( x\to z \right)\wedge \left( y\to z \right) \\ \end{align}$$

for all x, y, z ∈ H. Then (H,∨,∧,→,0,m) is called a Heyting ADL (HADL).

Let H be an HADL and → be a binary operation on H such that x →y ∈ H for any x, y ∈ H. Then y ≤ x →y implies that (x→y)∧y = y, but the converse is not always true. The converse becomes true whenever H is a lattice, and therefore an HADL becomes a Heyting algebra.

[3] Let H be an ADL with 0 and a maximal element m, then the following are equivalent.

(1)

H is an HADL

Definition 2.7. [4] Let P be a nonempty set and ≤,≤′ $\le ,{\le }'$be two partial orders on P. Then we say that ≤ and ≤′ ${\le }'$are dual to each other whenever for any a,b ∈ P, a ≤ b if and only if b≤′ $b{\le }'$a.

Note that the dual of any partial order ≤ on a nonempty set P is again a partial order on P and is unique. The poset (P , ≤′) $\left( P,{\le }' \right)$is the dual of the poset (P,≤).

Since the dual H ^d of a distributive lattice (H,∨,∧) is again a distributive lattice, we give the following definition. A distributive lattice (H,∨,∧) is called a dual Heyting algebra if its dual H ^d of H is a Heyting algebra.

[2] A distributive lattice (H,∨,∧,0,1) is a dual Heyting algebra if and only if there exists a binary operation ←satisfying the following:

(1)x←x=0(2)x∨(x←y)=x∨y(3)(x←y)∨y=y(4)x←(y∨z)=(x←y)∨(x←z)(5)(x∧y)←z=(x←z)∨(y←z)forallx,y,z∈H. $$\begin{align}& \left( 1 \right)x\leftarrow x=0 \\ & \left( 2 \right)x\vee \left( x\leftarrow y \right)=x\vee y \\ & \left( 3 \right)\left( x\leftarrow y \right)\vee y=y \\ & \left( 4 \right)x\leftarrow \left( y\vee z \right)=\left( x\leftarrow y \right)\vee \left( x\leftarrow z \right) \\ & \left( 5 \right)\left( x\wedge y \right)\leftarrow z=\left( x\leftarrow z \right)\vee \left( y\leftarrow z \right)for\,all\,x,y,z\,\in \,H. \\ \end{align}$$

Definition 2.9. [2] An ADL (H,∨,∧,0) with a maximal element m is called a dual Heyting almost distributive lattice (dual HADL), if for each a ∈ H the distributive lattice ([0,a],∨,∧,0,a) is a dual Heyting algebra with respect to the binary operation denoted by ←_a.

[2] Let H be an ADL with a maximal element m. Then the following are equivalent:

(1)

H is a dual HADL

(i)x←x=0(ii)((x←y)∨y)∧m=y∧m(iii)(x∨(x←y))∧m=(x∨y)∧m(iv)z←(x∨y)=(z←x)∨(z←y)(v)(x∧y)←z=(x←z)∨(y←z)forallx,y,z∈H. $$\begin{align}& \left( i \right)x\leftarrow x=0 \\ & \left( ii \right)\left( \left( x\leftarrow y \right)\vee y \right)\wedge m=y\wedge m \\ & \left( iii \right)\left( x\vee \left( x\leftarrow y \right) \right)\wedge m=\left( x\vee y \right)\wedge m \\ & \left( iv \right)z\leftarrow \left( x\vee y \right)=\left( z\leftarrow x \right)\vee \left( z\leftarrow y \right) \\ & \left( v \right)\left( x\wedge y \right)\leftarrow z=\left( x\leftarrow z \right)\vee \left( y\leftarrow z \right)\,for\,all\,x,y,z\,\in \,H. \\ \end{align}$$

Definition 2.11. [5] A skew lattice is an algebra S =(S,∧,∨) of type (2,2) such that ∧ and ∨ are both idempotent and associative, and they satisfy the following absorption laws:

x∧(x∨y) = x = x∨(x∧y) and (x∧y)∨y = y = (x∨y)∧y for all x, y ∈ S.

A skew lattice is called strongly distributive if for all x, y, z ∈ S it satisfies the following identities: x ∧(y∨z)= (x∧y)∨(x∧z) and (x∨y)∧z=(x∧z)∨(y∧z); and it is called co-strongly distributive if it satisfies the identities: x∨(y∧z) = (x∨y)∧(x∨z) and (x∧y)∨z = (x∨z)∧(y∨z).

Definition 2.12. [8] An algebra S = (S,∨,∧,→,1) of type (2,2,2,0) is said to be a skew Heyting algebra whenever the following conditions are satisfied:

A strongly distributive skew lattice (H,∨,∧) with bottom 0 is a dual skew Heyting algebra if and only if the following conditions are satisfied:

(1)

For any a ∈ H^d, where H^d is the dual of (H,∨,∧) an operation _a← can be defined on a↑ such that (a↑,∨,∧,0,a) is a dual Heyting algebra.

Through out this chapter for a non empty set L and for any a,b, c, z ∈ L we use the following notations:

(1)

L^z = {z∨c|c ∈ L}

For an ADL L and z ∈ L, the set L_z = {x∧z|x ∈ L} is a distributive lattice. But the set L^z is not necessarily a distributive lattice. This motivates us to study an algebra defined in terms of the set L^z.

Definition 3.3. Let L be an ADL with out a maximal element m but with 0. Then L is said to be a dual skew HADL if, for each z ∈ L the algebra (L^z,∨,∧, _z←, z) is a dual skew Heyting algebra.

Example 3.4. Let L be an ADL with 0. For any z ∈ L we define a binary operation←_z on L_z by

x←zy={ 0ifx≥yyotherwise. $$x{{\leftarrow }_{z}}y=\left\{ \begin{matrix}0 & \text{if}\,x\ge y \\y & \text{otherwise}\text{.} \\\end{matrix} \right.$$

It is easy to show that (L_z,∨,∧,←_z,0, z) is a dual Heyting algebra.

The following theorem characterises dual skew HADLs.

Let L be an ADL with 0. Then the following conditions are equivalent.

(1)

L is a dual skew HADL

Proof. Assume L to be a dual skew HADL. This implies that for any z ∈ L, L^z is a dual skew Heyting algebra and hence it is a skew lattice. In particular L⁰ = L is a dual skew Heyting algebra. Consequently, from the definition of dual skew Heyting algebra we have seen that for any b ∈ L there exists z ∈ L^z and ([z,b],∨,∧,←_b, z,b) is a dual Heyting algebra. Hence [z,b] is a dual HADL. Since L^z is a dual skew Heyting algebra, the induced operation _z←on L^z from←_a on [z,a], is given by x_z←y = (y∧x∧y)←_y y. Thus it is possible to define a binary operation ←on L by x←y = (y∧x∧y)←_y y = (x∧y)←_y y.

Conversely, suppose that condition (2) hold and let z ∈ L. Since the meet operation in an ADL is distributive over the join operation, and by (i), L^z is a strongly distributive skew lattice with bottom z. By (ii) for any b ∈ L^z, [z,b] is a dual HADL. Since [z,b] is a lattice ([z,b],∨,∧,←_b, z,b) is a dual Heyting algebra. Now using (ii) it is possible to define _z←on L^z by x _z←y = (x∧y)←_y y. But x∧y = y∧x∧y. Hence x _z←y = (y∧x∧y)←_y y. Therefore (L^z,∨,∧, _z←, z) is a dual skew Heyting algebra and hence L is a dual skew HADL.

Corollary 3.6. Let L be a dual skew HADL. Then for any a ∈ L,L_a is a dual Heyting algebra.

Proof. Clear by Theorem 3.5(2).

Let L be a dual skew HADL. Then for any z ∈ L,L^z is a dual skew HADL.

Proof. Suppose that L is a dual skew HADL. Then for any z ∈ L, L^z is a dual skew Heyting algebra. Take any y ∈ L^z as y ∈ L, L^y is also a dual skew Heyting algebra. Therefore L^z is a dual skew HADL.

Here is an important result that we use in the sequel.

Corollary 3.8. Let L be a dual skew HADL. If x, y ∈ L such that x ≤ y and a,b ∈ L_x, then a _x← b = a _y← b.

Proof. Let L be a dual skew HADL and x, y ∈ L such that x ≤ y. Then L_x ⊆ L_y. If a,b ∈ L_x, then a←_x b ∈ L_x and hence a←_x b ∈ L_y. Since a,b ∈ L_y, a←_y b also belongs to L_y. The minimal element characterisation of a←_x b and a←_y b on the dual Heyting algebra L_y forces the two elements to be equal.

The next theorem characterises a dual skew HADL interms of a congruence relation θ defined on it. First observe the following lemma.

Let H be a dual skew HADL and x, y, z ∈ H such that x ∧m = y ∧m. Then the following statements hold:

(1)x←y=0(2)x←z=y←z. $$\begin{align}& \left( 1 \right)x\leftarrow y=0 \\ & \left( 2 \right)x\leftarrow z=y\leftarrow z. \\ \end{align}$$

Proof. Let L be a dual skew HADL and x, y, z ∈ L. Clearly, x ∧ y = y and y ∧ x = x. Then (1) x ← y = x ← (x∧y) = (x∧(x∧y))←_(x∧y) (x∧y) = (x∧y)←_(x∧y) (x∧y) = 0.

To prove (2), taking z ∈ L, we obtain that

x←z=(x∧z)←zz=((y∧x)∧z)←zz=((x∧y)∧z)←zz=(y∧z)←zz=y←z. $$\begin{align}& x\leftarrow z=\left( x\wedge z \right){{\leftarrow }_{z}}z \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( y\wedge x \right)\wedge z \right){{\leftarrow }_{z}}z \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( x\wedge y \right)\wedge z \right){{\leftarrow }_{z}}z \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( y\wedge z \right){{\leftarrow }_{z}}z \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,=y\leftarrow z. \\ \end{align}$$

Let L be a relatively complemented ADL with 0. If L is a dual skew HADL and θ is defined by

θ={ (x,y)∈L×L|x∨y=xandy∨x=y }, $$\theta =\left\{ \left( x,y \right)\in L\times L|x\vee y=x\,and\,y\vee x=y \right\},$$

then θ is an equivalence relation on L.

Proof. Suppose L is a dual skew HADL. Let x, y, z ∈ L. It is easly seen that θ is reflexive and symmetric. Assume that (x, y) ∈ θ and (y, z) ∈ θ. Then x∨y = x, y∨x = y, y∨z = y and z∨y = z. Hence z∨x = (z∨y)∨x = z∨(y∨x) = z∨y = z and x∨z = (x∨y)∨z = x∨(y∨z) = (x∨y) = x, it follows that (x,z) ∈ θ. Consequently θ is transitive and therefore it is an equivalence relation.

Let a relatively complemented ADL L with 0 be a dual skew HADL. For each b ∈ L, consider the equivalence class [b]θ where θ is a relation in the above lemma. Then the following conditions hold:

(1)

θ is a congruence relation on [b]θ

Proof. To show that θ is a congruence relation on [b]θ, for any x, y,a ∈ [b]θ we need to check that ((x←a), (y← a)) ∈ θ, ((a ← x), (a ← y)) ∈ θ and the substitution properties are satisfied. Clearly for any x, y,a ∈ [b]θ, (x∧a)∨(y∧a) = (x∨y)∧a = x∧a and (y∧a)∨(x∧a) = (y∨x)∧a = y∧a. Thus ((x∧a), (y∧a)) ∈ θ.

Moreover,

(x∨a)∧(y∨a)=((x∨a)∧y)∨((x∨a)∧a)=((x∨a)∧y)∨a=((a∨x)∧y)∨a=((a∧y)∨(x∧y))∨a=((a∧y)∨y)∨a=y∨a. $$\begin{align}& \left( x\vee a \right)\wedge \left( y\vee a \right)=\left( \left( x\vee a \right)\wedge y \right)\vee \left( \left( x\vee a \right)\wedge a \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( x\vee a \right)\wedge y \right)\vee a \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( a\vee x \right)\wedge y \right)\vee a \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( a\wedge y \right)\vee \left( x\wedge y \right) \right)\vee a \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( a\wedge y \right)\vee y \right)\vee a \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=y\vee a. \\ \end{align}$$

Using the same procedure we have (y ∨ a) ∧ (x ∨ a) = x ∨ a. From these results we obtain that (x ∨ a) ∨ (y∨a) = (x∨a)∨((x∨a)∧(y∨a)) = x∨a and (y∨a)∨(x∨a) = (y∨a)∨((y∨a)∧(x∨a)) = y∨a. Thus ((x∨a), (y∨a)) ∈ θ and ((y∨a), (x∨a)) ∈ θ. This shows that θ satisfies the substitution property.

One can simply observe that x ∧y = (x∨y)∧y = y and y ∧x = (y∨x)∧x = x. Indeed, point (2) of the above lemma assures that ((x←a), (y←a)) ∈ θ. Furthermore, using Lemma 3.9 (1) we obtain that (a←x)∨(a← y) = (a←x)∨0 = a←x and (a←y)∨(a←x) = (a←y)∨0 = a←y. Thus ((a←x), (a←y)) ∈ θ. Hence θ is a congruence relation on each equivalence class.

Suppose y, z ∈ [b]θ. It is obvious that (y, z) ∈ θ. Following this we obtain that z∧y = (z∨y)∧(y∨z) = (y ∨ z) ∧ (y ∨ z) = y ∨ z so that each congruence class is rectangular. Now take an arbitrary element x ∈ [b]θ and consider the congruence class [x]θ. Let T be a rectangular subalgebra of [b]θ such that [x]θ ⊆ T. Let r ∈ T. Since x ∈ T and T is a rectangular subalgebra of [x]θ we have r ∨ x = x ∧ r and x ∨ r = r ∧ x. Thus x∧r = (r∧x)∧r = (x∨r)∧r = r which implies that r∨x = x∧r = r and x∨r = x∨(x∧r) = x. Hence r ∈ [x]θ. Therefore T ⊆ [x]θ and we conclude that [x]θ = T. Hence [x]θ is maximal, i.e., each congruence class [x]θ is a maximal rectangular subalgebra of [b]θ. Next, to show that [x]θ/θ is a lattice we need to have the following results. Let x, y ∈ [x]θ. Then

x∨y=((y∨x)∧x)∨y=((y∨x)∧x)∨((x∨y)∧y)=((y∨x)∧x)∨((y∨x)∧y)=(y∨x)∧(x∨y). $$\begin{align}& x\vee y=\left( \left( y\vee x \right)\wedge x \right)\vee y \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( y\vee x \right)\wedge x \right)\vee \left( \left( x\vee y \right)\wedge y \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( y\vee x \right)\wedge x \right)\vee \left( \left( y\vee x \right)\wedge y \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\left( y\vee x \right)\wedge \left( x\vee y \right). \\ \end{align}$$

Similarly,

y∨x=((x∨y)∧y)∨x=((x∨y)∧y)∨((y∨x)∧x)=((x∨y)∧y)∨((x∨y)∧x)=(x∨y)∧(y∨x). $$\begin{align}& y\vee x=\left( \left( x\vee y \right)\wedge y \right)\vee x \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( x\vee y \right)\wedge y \right)\vee \left( \left( y\vee x \right)\wedge x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( x\vee y \right)\wedge y \right)\vee \left( \left( x\vee y \right)\wedge x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,=\left( x\vee y \right)\wedge \left( y\vee x \right). \\ \end{align}$$

Hence (x∨y)∨(y∨x) = (x∨y)∨((x∨y)∧(y∨x))=x∨y and (y∨x)∨(x∨y) = (y∨x)∨((y∨x)∧(x∨y))=y∨x.

Therefore ((x∨y), (y∨x)) ∈ θ.

Similarly, we have

(x∧y)∨(y∧x)=(x∧y)∨(x∧y∧x)=(x∧y)∨((x∧y)∧x)=x∧y $$\begin{align}& \left( x\wedge y \right)\vee \left( y\wedge x \right)=\left( x\wedge y \right)\vee \left( x\wedge y\wedge x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( x\wedge y \right)\vee \left( \left( x\wedge y \right)\wedge x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=x\wedge y \\ \end{align}$$

and

(y∧x)∨(x∧y)=(y∧x)∨(y∧x∧y)=(y∧x)∨((y∧x)∧y)=y∧x. $$\begin{array}{l}{(y\wedge x)}\vee{(x\wedge y)}={(y\wedge x)}\vee{(y\wedge x\wedge y)}\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }={(y\wedge x)}\vee{({(y\wedge x)}\wedge y)}\\\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }\text{ }=y\wedge x.\end{array}$$

Hence, ((x∧y), (y∧x)) ∈ θ.

Now, assume that [y]θ, [z]θ ∈ [x]θ/θ. Define [y]θ ∨ [z]θ = [y ∨ z]θ and [y]θ ∧ [z]θ = [y ∧ z]θ. Then t ∈ [y∨z]θ ⇔(t, (y∨z)) ∈ θ ⇔(t, (z∨y)) ∈ θ ⇔t ∈ [z∨y]θ. Thus [y]θ ∨[z]θ = [z]θ ∨[y]θ. Similarly, [y]θ ∧[z]θ = [z]θ ∧[y]θ. Therefore [x]θ/θ is a lattice. Let β be a congruence relation on [x]θ such that [x]θ/β is a lattice. Suppose (y, z) ∈ θ. Clearly

y∨z=yandz∨y=z⇒[ y∨z ]β=[ y ]βand[ z∨y ]β=[ z ]β⇒[ y ]β∨[ z ]β=[ y ]βand[ z ]β∨[ y ]β=[ z ]β⇒[ y ]β=[ z ]β,since[ y ]β∨[ z ]β=[ z ]β∨[ y ]β⇒(y,z)∈β. $$\begin{align}& y\vee z=y\,\text{and}\,z\vee y=z\Rightarrow \left[ y\vee z \right]\beta =\left[ y \right]\beta \,\text{and}\,\left[ z\vee y \right]\beta =\left[ z \right]\beta \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \left[ y \right]\beta \vee \left[ z \right]\beta =\left[ y \right]\beta \,\text{and}\,\left[ z \right]\beta \,\vee \left[ y \right]\beta =\left[ z \right]\beta \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \left[ y \right]\beta =\left[ z \right]\beta ,\,\text{since}\left[ y \right]\beta \,\vee \,\left[ z \right]\beta =\left[ z \right]\beta \vee \left[ y \right]\beta \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Rightarrow \left( y,z \right)\in \beta . \\ \end{align}$$

Therefore, θ ⊆ β. Suppose H is a lattice image of L. Then there exist an epimorphism f : [x]θ −→ H. Define the kernel K of f by

K={ (y,z)∈[ b ]θ2|f(y)=f(z) }. $$K=\left\{ \left( y,z \right)\in \left[ b \right]{{\theta }^{2}}|f\left( y \right)=f\left( z \right) \right\}.$$

K is reflexive, symmetric and transitive. Let (x₁, y₁), (x₂, y₂) ∈ K. Since f is a homomorphism, we have f (x₁ ∨ x₂) = f (x₁)∨ f (x₂) = f (y₁)∨ f (y₂) = f (y₁ ∨y₂) so that ((x₁ ∨x₂), (y₁ ∨y₂)) ∈ K. Using a similar procedure for ∧ and ← we obtain that ((x₁ ∧x₂), (y₁ ∧y₂)) ∈ K and ((x₁ ← x₂), (y₁ ← y₂)) ∈ K. This shows that K is a congruence relation on [x]θ. Hence we obtain that [x]θ/Ker f f ([x]θ) = H. Therefore any lattice image of [x]θ is of the form [x]θ/ϕ for some congruence relation ϕ on [x]θ.

Consider that the set L = {[x]θ/ϕ|ϕ is a congruence on [x]θ and θ ⊆ ϕ } of lattices. Now, take the congruence relations θ and β discussed earlier and assume that [x]θ/θ ⊆ [x]θ/β. Let a, c ∈ [x]θ such that (a, c) ∈ β. Since θ ⊆ β and [x]θ/θ ⊆ [x]θ/β there exist a′, c′∈ [x]θ such that a ∈ [a]θ = [a′]β and c ∈ [x]θ = [c′]β. Thus (a,a′), (c, c′) ∈ θ and (a,a′), (c, c′) ∈ β. Hence (a, c), (c, c′) ∈ β ⇒(a, c′) ∈ β ⇒a ∈ [b′]β = [x]θ ⇒(a, c) ∈ θ. Therefore β ⊆ θ and hence θ = β. Consequently, [x]θ/θ = [x]θ/β so that [x]θ/θ is the maximal lattice image of [x]θ.

Let L be an ADL with 0 and z ∈ L. If L is a dual skew HADL, then for any w, x, y ∈ L_z the following conditions hold:

(1)

x ≥ w←_z x

Proof. Suppose that L is a dual skew HADL and z ∈ L such that w, x, y ∈ L_z. Then by definition L^z is a dual skew Heyting algebra. This implies that L⁰ = L is a dual skew Heyting algebra so that L_z is a dual Heyting algebra. Thus (1) follows. Now, assume that (x←_z y)∨w = w. Then

w∨x=((x←zy)∨w)∨x=((x←zy)∨x)∨w=x∨y∨w=y∨w∨x. $$\begin{align}& w\vee x=\left( \left( x{{\leftarrow }_{z}}y \right)\vee w \right)\vee x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( x{{\leftarrow }_{z}}y \right)\vee x \right)\vee w \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=x\vee y\vee w \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,=y\vee w\vee x. \\ \end{align}$$

On the other hand given that y∨w∨x = w ∨x, we obtain x ←_z (w∨x) = x ←_z (y∨w∨x). Hence (x←_z w) = (x←_z y)∨(x←_z w). Thus

(x←zy)∨w=(x←zy)∨((x←zw)∨w)=((x←zy)∨(x←zw))∨w=(x←zw)∨w=w. $$\begin{align}& \left( x{{\leftarrow }_{z}}y \right)\vee w=\left( x{{\leftarrow }_{z}}y \right)\vee \left( \left( x{{\leftarrow }_{z}}w \right)\vee w \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( x{{\leftarrow }_{z}}y \right)\vee \left( x{{\leftarrow }_{z}}w \right) \right)\vee w \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( x{{\leftarrow }_{z}}w \right)\vee w \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=w. \\ \end{align}$$

Therefore (2) is satisfied.

Let L be an ADL with 0 and for any z ∈ L, L^z is a skew lattice. Then L is a dual skew HADL if for any b ∈ L^z and w, x, y ∈ [z,b] the following conditions are satisfied:

(1)

x ≥ w←_b x

Proof. Assume that conditions (1) and (2) hold. For any b ∈ L^z and w, x, y ∈ [z,b] we have by (1), y ≥ x ←_b y such that x ←_b y ∈ [z,b] and then by (2) we get

w≥x←by⇔w=(x←by)∨w⇔w∨x=y∨w∨x⇔w∨x≥y. $$\begin{align}& w\ge x{{\leftarrow }_{b}}y\Leftrightarrow w=\left( x{{\leftarrow }_{b}}y \right)\vee w \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow w\vee x=y\vee w\vee x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\Leftrightarrow w\vee x\ge y. \\ \end{align}$$

Hence, [z,b] is a dual Heyting algebra. Defining a binary operation _z←on L^z by x_z←y = (y∧x∧y)←_y y makes L^z is a dual skew Heyting algebra and therefore L is a dual skew HADL.

Corollary 3.14. On a dual skew HADL L, (x∧y)←_y y = 0 if and only if x ∨y = x.

Proof. Suppose that L is a dual skew HADL and (x∧y)←_y y = 0. Since L_y is a dual Heyting algebra (Corollary 3.6), we obtain that

x = x ∨<!-- ∨ --> x ∧<!-- ∧ --> y = x ∨<!-- ∨ --> x ∧<!-- ∧ --> y ∨<!-- ∨ --> 0 = x ∨<!-- ∨ --> x ∧<!-- ∧ --> y ∨<!-- ∨ --> x ∧<!-- ∧ --> y ←<!-- ← --> y y = x ∨<!-- ∨ --> x ∧<!-- ∧ --> y ∨<!-- ∨ --> y = x ∨<!-- ∨ --> y . $$\begin{align} & \,\,\,\,\,x=x\vee \left( x\wedge y \right) \\ & \,\,\,\,\,\,\,\,\,=x\vee \left( \left( x\wedge y \right)\vee 0 \right) \\ & \,\,\,\,\,\,\,\,\,=x\vee \left( \left( x\wedge y \right)\vee \left( \left( x\wedge y \right){{\leftarrow }_{y}}y \right) \right) \\ & \,\,\,\,\,\,\,\,=x\vee \left( \left( x\wedge y \right)\vee y \right) \\ & \,\,\,\,\,\,\,\,=x\vee y. \\ \end{align}$$

Hence x ∨y = x. The converse is straight forward.

Let L is a skew ADL with 0. If the set PI (L) of all principal ideals of L is a dual skew Heyting algebra, then L is a dual skew HADL.

Proof. Suppose PI (L) be a dual skew Heyting algebra. For each x ∈ L we show that L^x is a dual skew Heyting algebra. Define _x← on L^x by a _x← b = c∧x, for some c ∈ L such that (a]←(b] = (c]. Let (u] = (v] for some u, v ∈ L. Since (u]⊆(v] and (v]⊆(u] implies that u∧v=v and v ∧u=u respectively, we have u∧x =v∧u∧x = u∧v∧x = v∧x. Now take a,b, c,d ∈ L^x such that a = c,b = d. Then we obtain that (a] = (c] and (b] = (d]. Consequently (a] ← (b] = (c] ← (d]. Let (a] ← (b] = (e], (c] ← (d] = (f ] for some e, f ∈ L, we have a _x← b = e ∧x and c _x←d = f ∧x. Hence e ∧x = f ∧x implies that a _x←b = c _x←d. Therefore the binary operation _x← is well defined on L^x. It is clear that L^x is a strongly distributive skew lattice with bottom x. For any y ∈ L^x, we claim that [x, y] is a dual Heyting algebra. Let a,b, c ∈ [x, y] and define ←_y on [x, y] by a ←_y b = a _x← b. Clearly (a], (b], (c] ∈ (y]↓ and (y]↓ is a dual Heyting algebra. Then

(i)

(a]←(a] = (y]. Then we have a←_y a = a←_x a = y∧x = x.

(a←yb)∨b=(ax←b)∨b=(t∧x)∨b=(b∧t∧x)∨b=(t∧b∧x)∨b=(t∧x∧b)∨b=b. $$\begin{align}& \left( a{{\leftarrow }_{y}}b \right)\vee b=\left( a{{\,}_{x}}\leftarrow b \right)\vee b \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( t\wedge x \right)\vee b \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( b\wedge t\wedge x \right)\vee b \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( t\wedge b\wedge x \right)\vee b \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( t\wedge x\wedge b \right)\vee b \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=b. \\ \end{align}$$

This is because of (a]←(b] ⊆ (b]⇒(t] ⊆ (b]⇒b∧t = t.

(iii)

Let (a] ← (b] = (t] for some t ∈ L. Then a _x← b = t ∧x. Now (a∨t] = (a]∨(t] = (a]∨((a] ← (b]) = (a]∨(b] = (a∨b]. Hence (a∨t)∧x = (a∨b)∧x. Therefore

a∨(a←yb)=a∨(ax←b)=a∨(t∧x)=(a∨t)∧(a∨x)=(a∨t)∧x=(a∨b)∧x=a∨b. $$\begin{align}& a\vee \left( a{{\leftarrow }_{y}}b \right)=a\vee \left( a{{\,}_{x}}\leftarrow b \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=a\vee \left( t\wedge x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( a\vee t \right)\wedge \left( a\vee x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( a\vee t \right)\wedge x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( a\vee b \right)\wedge x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=a\vee b. \\ \end{align}$$

(iv)

Let (a]←(c] = (u] and (b]←(c] = (v] for some u, v ∈ L. Then we obtain that a _x← c = u∧x and b _x← c = v ∧x. Further

( a∧b ]←( c ]=(( a ]∧( b ])←( c ]=(( a ]←( c ])∨(( b ]←( c ])=( u ]∨( v ]=( u∨v ]. $$\begin{align}& \left( a\wedge b \right]\leftarrow \left( c \right]=\left( \left( a \right]\wedge \left( b \right] \right)\leftarrow \left( c \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( a \right]\leftarrow \left( c \right] \right)\vee \left( \left( b \right]\leftarrow \left( c \right] \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u \right]\vee \left( v \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u\vee v \right]. \\ \end{align}$$

Hence we obtain that

(a∧b)←yc=(a∧b)x←c=(u∨v)∧x=(u∧x)∨(v∧x)=(ax←c)∨(bx←c)=(a←yc)∨(b←yc). $$\begin{align}& \left( a\wedge b \right){{\leftarrow }_{y}}c=\left( a\wedge b \right){{\,}_{x}}\leftarrow c \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u\vee v \right)\wedge x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u\wedge x \right)\vee \left( v\wedge x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( a{{\,}_{x}}\leftarrow c \right)\vee \left( b{{\,}_{x}}\leftarrow c \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( a{{\leftarrow }_{y}}c \right)\vee \left( b{{\leftarrow }_{y}}c \right). \\ \end{align}$$

(v)

Let (a]←(b] = (u] and (a]←(c] = (v] for some u, v ∈ L. Then a _x← b = u∧x and a _x← c = v ∧x. Since

a ←<!-- ← --> b ∨<!-- ∨ --> c = a ←<!-- ← --> b ∨<!-- ∨ --> c = a ←<!-- ← --> b ∨<!-- ∨ --> a ←<!-- ← --> c = u ∨<!-- ∨ --> v = u ∨<!-- ∨ --> v , $$\begin{align}& \left( a \right]\leftarrow \left( b\vee c \right]=\left( a \right]\leftarrow \left( \left( b \right]\vee \left( c \right] \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( a \right]\leftarrow \left( b \right] \right)\vee \left( \left( a \right]\leftarrow \left( c \right] \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u \right]\vee \left( v \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u\vee v \right], \\ \end{align}$$

we get

a ←<!-- ← --> y b ∨<!-- ∨ --> c = a x ←<!-- ← --> b ∨<!-- ∨ --> c = u ∨<!-- ∨ --> v ∧<!-- ∧ --> x = u ∧<!-- ∧ --> x ∨<!-- ∨ --> v ∧<!-- ∧ --> x = a x ←<!-- ← --> b ∨<!-- ∨ --> a x ←<!-- ← --> c = a ←<!-- ← --> y b ∨<!-- ∨ --> a ←<!-- ← --> y c . $$\begin{align}& a{{\leftarrow }_{y}}\left( b\vee c \right)=a{{\,}_{x}}\leftarrow \left( b\vee c \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u\vee v \right)\wedge x \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( u\wedge x \right)\vee \left( v\wedge x \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( a{{\,}_{x}}\leftarrow b \right)\vee \left( a{{\,}_{x}}\leftarrow c \right) \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( a\,{{\leftarrow }_{y}}b \right)\vee \left( a{{\leftarrow }_{y}}c \right). \\ \end{align}$$

Hence [x, y] is a dual Heyting algebra.

On PI (L) we have (a]←_(b] (b] = (a←_b b]. Then for a,b ∈ L^x we have the following

( a ]←( b ]=(( b ]∧( a ]∧( b ])←( b ]( b ]=( b∧a∧b ]←( b ]( b ]=( (b∧a∧b)←bb ]. $$\begin{align}& \left( a \right]\leftarrow \left( b \right]=\left( \left( b \right]\wedge \left( a \right]\wedge \left( b \right] \right){{\leftarrow }_{\left( b \right]}}\left( b \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( b\wedge a\wedge b \right]{{\leftarrow }_{\left( b \right]}}\left( b \right] \\ & \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\left( \left( b\wedge a\wedge b \right){{\leftarrow }_{b}}b \right]. \\ \end{align}$$

This implies that a _x← b = ((b∧a∧b)←_b b)∧x. Thus a←_x b = (b∧a∧b)←_b b. Hence L^x is a dual skew Heyting algebra. Therefore L is a dual skew HADL.

Corollary 3.16. On a dual skew HADL H, (x∧y)_y←y = 0 if and only if x ∨y = x.

Proof. Suppose that H is a dual skew HADL and (x∧y)_y←y = 0. Then

x = x ∨<!-- ∨ --> 0 = x ∨<!-- ∨ --> x ∧<!-- ∧ --> y y ←<!-- ← --> y = x ∨<!-- ∨ --> x y ←<!-- ← --> y ∨<!-- ∨ --> y y ←<!-- ← --> y = x ∨<!-- ∨ --> x y ←<!-- ← --> y ∨<!-- ∨ --> 0 = x ∨<!-- ∨ --> x y ←<!-- ← --> y = x ∨<!-- ∨ --> y . $$\begin{align}& x=x\vee 0 \\ & \,\,\,\,=x\vee \left( {{\left( x\wedge y \right)}_{y}}\leftarrow y \right) \\ & \,\,\,\,=x\vee \left( \left( {{x}_{y}}\leftarrow y \right)\vee \left( {{y}_{y}}\leftarrow y \right) \right) \\ & \,\,\,\,=x\vee \left( {{x}_{y}}\leftarrow y \right)\vee 0 \\ & \,\,\,\,=x\vee \left( {{x}_{y}}\leftarrow y \right) \\ & \,\,\,\,=x\vee y. \\ \end{align}$$

Hence x ∨y = x. The converse is straight forward.