Definition 3.1. *Let A be a non-void subset of a BL-algebra L, then we call the set A*^{⊥}= {*x* ∈ *L* | *a* ∧ *x* = 0 *for all a* ∈ *A*} *is an annihilator of A*.

*Let L* ={0*, a, b, c, d*, 1} *be a set, where* 0 ≤ *a* ≤ *b* ≤ 1, 0 ≤ *a* ≤ *d* ≤ 1 *and* 0 ≤ *c* ≤ *d* ≤ 1*. The Cayley tables are as follows*:

*Then* (*L*, ∧, ∨, ⊙, →, 0, 1) *is a BL-algebra. Now, we consider A* = {0*, a*}*, it is easy to check that A*^{⊥} = {0, *c*}.

*Let L be a BL-algebra and A be a subset of L, then A*^{⊥} *is an ideal of L. Moreover, if A* ≠ {0}*, then A*^{⊥} *is proper*.

*Proof*. For every *a* ∈ *A*, we have *a* ∧ 0 = 0, hence 0 ∈ *A*^{⊥}, which implies *A*^{⊥} is nonempty.

I1: Assume that *y* ∈ *A*^{⊥}, *x* ≤ *y*, since for all *a* ∈ *A*, *x* ∧ *a* ≤ *y* ∧ *a* = 0, then we have *x* ∈ *A*^{⊥};

I2: Assume that *x* ∈ *A*^{⊥}, *y* ∈ *A*^{⊥}. Let *a* ∈ *A*, then *a* ∧ (*x* → *y*)=*a* ⊙ (*a* → (*x* → *y*))=*a* ⊙ (*x* → (*a* → *y*)) ≤ *a* ⊙ (*x* ⊙ *a* → *a* ⊙ (*a* → *y*))=*a* ⊙ ((*x* ⊙ *a*) → (*a* ∧ *y*)) = *a* ⊙ ((*x* ⊙ *a*) → 0)=*a* ⊙ (*x* → *a*) ≤ *a* ⊙ $(\overline{\overline{a}}\to $$\overline{\overline{x}}=\overline{\overline{a}}\wedge \overline{\overline{x}}=\overline{\overline{a\wedge x}}=0$. Hence *x* ⊘ *y* = *x* → *y* ∈ *A*^{⊥}, which implies *A*^{⊥} is an ideal.

If *A* ≠ {0}, then there is *a* ∈ *A* such that *a* ≠ 0, so 1 ∧ *a* = *a* ≠ 0, then we have 1 ∉ *A*^{⊥}. Therefore, *A*^{⊥} is proper. □

*Let L be a BL-algebra. Then the following conclusions hold: for all x, a, b* ∈ *L*,

{1}^{⊥} = {0}, {0}^{⊥} = *L;*

*if a* ≤ *b, then* {*b*}^{⊥} ⊆ {*a*}^{⊥};

{*a*}^{ ⊥} ∩ {*b*}^{⊥} = {*a* ∨ *b*}^{⊥};

{*a*}^{⊥} ∪{*b*}^{⊥} ⊆ { *a* ∧ *b*}^{⊥};

*if x* ∈ {*a*}^{⊥}*, then a* ∨ *x* *and x* ≤ *a*.

*Proof*. (1) For all *x* ∈ {1}^{⊥}, *x* = *x* ∧ 1 = 0, so *x* = 0, which implies {1}^{⊥} ={0}. For all *x* ∈ *L*, since *x* ∧ 0 = 0, we have *L* ⊆ {0}^{⊥}, and evidently, {0}^{⊥} ⊆ *L*, so {0}^{ ⊥} = *L*.

(2) For all *x* ∈ {b}^{⊥}, we have *a* ∧ *x* ≤ *b* ∧ *x* = 0, so *a* ∧ *x* = 0, so we have *x* ∈ {a}^{⊥}.

(3) *x* ∈ {a}^{⊥} ∩ {b}^{⊥} iff *x* ∈ {a}^{⊥} and *x* ∈ {b}^{⊥} iff *x* ∧ *a* = 0 and *x* ∧ *b* = 0 iff *x* ∧ (*a* ∨ *b*) = 0 iff *x* ∈ {*a* ∨ *b*}^{⊥}.

(4) If *x* ∈ {a}^{⊥}[{b}^{⊥}, then *x* ∈ {a}^{⊥} or *x* ∈ {b}^{⊥}, so *x* ∧ *a* = 0 or *x* ∧ *b* = 0, and so *x* ∧ *a* ∧ *b* = 0, therefore, *x* ∈ {*a* ∧ *b*}^{⊥}.

(5) If *x* ∈ {a}^{⊥}, then *x* ∧ *a* = 0, so we have *a* ⊙ *x* ≤ *x* ∧ *a* = 0, then *a* ⊙ *x* = 0, therefore, *a* ∨ *x* and *x* ≤ *a*. □

*Let L be the BL-algebra in Example* 3.2*. We have a*^{⊥} = {0, *c*}*, c*^{⊥} = {0*, a, b*}*, and a* ∧ *c* = 0*, so a*^{⊥} [*c*^{⊥} = {0*, a, b, c*}*. Hence* 0^{⊥} = *L* ⊈ *a*^{⊥} [*c*^{⊥}*. Therefore, we do not have the equation for Proposition 3.4(4)*.

*Let L be a BL-algebra. Then the following conclusions hold: for all x, y, a, b* ∈ *L*,

*if x* ∈ {a}^{⊥}*, y* ∈ {*a* ∨ *b*}^{⊥}*, then x* ∧ *y* ∈ {*a* ∧ *b*}^{⊥};

*if x* ∈ {a}^{⊥}*, y* ∈ {*a* → *b*}^{⊥}*, then x* ∧ *y* ∈ {*a* ∧ *b*}^{⊥};

*if x* ∈ {a}^{⊥}*, y* ∈ {b}^{⊥}*, then x* ⊙ *y, x* ∨ *y, x* ∧ *y* ∈ {*a* ∧ *b*}^{⊥}.

*Proof*. (1) Since *y* ∈ {*a* ∨ *b*}^{⊥} and *b* ≤ *a* ∨ *b*, by Proposition 3.4 (2)*, we have* {*a* ∨ *b*}^{⊥} ⊆ {b}^{⊥}, so *y* ∈ {b}^{⊥},

since {b}^{⊥} is a down set, we get *x* ∧ *y* ∈ {b}^{⊥} ⊆ {*a* ∧ *b*}^{⊥}.

(2) Since *y* ∈ {a → *b*}^{⊥} and *b* ≤ *a* → *b*, by Proposition 3.4 (2)*, we get* {a → *b*}^{⊥} ⊆ {b}^{⊥}, so *y* ∈ {b}^{⊥}. Since {b}^{⊥} is a down set, we get *x* ∧ *y* ∈ {b}^{⊥} ⊆ {a ∧ *b*}^{⊥}.

(3) Since *x* ∈ {a}^{⊥}, *y* ∈ {b}^{⊥} and *a* ∧ *b* ≤ *a, b*, by Proposition 3.4 (2)*, we have* {a}^{⊥} ⊆ {a ∧ *b*}^{⊥}, {b}^{⊥} ⊆ {a ∧ *b*}, so *x, y* ∈ {a ∧ *b*}^{⊥}, and so *x* ∨ *y* ∈ {a ∧ *b*}^{⊥}. Moreover, since {a ∧ *b*}^{⊥} is a down set, we get *x* ⊙ *y, x* ∧ *y* ∈ {a ∧ *b*}. □

*For any* ∅ ≠ *X* ⊆ *L*, 〈*X*〉∩ *X*^{⊥} = {0}.

*Proof*. Assume that *a* ∈ 〈*X*〉 ∩ *X*^{⊥}, then we have *a* ≤ *x*_{1} ⊘ *x*_{2} ⊘ · · · ⊘ *x*_{n} x_{i} ∈ *X*, and *a* ∧ *x*_{i} = 0 *i* = 1, 2, · · · *n*.

Now we prove that *a* ∧(*x*_{1} ⊘*x*_{2} ⊘· · · ⊘*x*_{n}) ≤ (*a* ∧*x*_{1}) ⊘(*a* ∧*x*_{2}) ⊘· · · ⊘(*a* ∧*x*_{n}). Firstly, *a* ∧(*x* ⊘*y*) → ((*a*∧*x*)⊘(*a*∧*y*))≤*a*∧*x* → ((*a* ∧ x) → (*a*∧*y*)) = ((*a*∧*x*) ⊙ (*a* ∧ x)) → (*a*∧*y*) = 0 → (*a*∧*y*) = 1, so *a*∧(*x* ⊘*y*) ≤ (*a* ∧*x*) ⊘(*a* ∧*y*). Then assume that when *k* = *n*, *a* ∧ (*x*_{1} ⊘*x*_{2} ⊘· · · ⊘*x*_{n}) ≤ (*a* ∧ *x*_{1}) ⊘(*a* ∧*x*_{2}) ⊘· · · ⊘(*a* ∧*x*_{n}) holds. When *k* = *n* + 1, *a* ∧ (*x*_{1} ⊘ *x*_{2} ⊘· · · ⊘ *x*_{n} ⊘ *xn*+1) ≤ (*a* ∧ (*x*_{1} ⊘ *x*_{2} ⊘· · · ⊘ *x*_{n}) ⊘ (*a* ∧ *xn*+1) ∨ (*a* ∧ *x*_{1}) ⊘ (*a* ∧ *x*_{2}) ⊘· · · ⊘ (*a* ∧ *x*_{n}) ⊘ (*a* ∧ *xn*+1).

Considering the above we have *a* ∧ (*x*_{1} ⊘ *x*_{2} ⊘· · · ⊘ *x*_{n}) ≤ (*a* ∧ *x*_{1}) ⊘ (*a* ∧ *x*_{2}) ⊘· · · ⊘ (*a* ∧ *x*_{n}). Then *a* ∧ *a* ≤ *a* ∧ (*x*_{1} ⊘ *x*_{2} ⊘· · · ⊘ *x*_{n}) ≤ (*a* ∧ *x*_{1}) ⊘ (*a* ∧ *x*_{2}) ⊘· · · ⊘ (*a* ∧ *x*_{n)} = 0 ⊘ 0 ⊘· · · ⊘ 0 = 0. Therefore, 〈*X*〉∩ *X*^{ ⊥} = {0}. □

*Let A be an ideal of L and A be linear*(*which means that A is totally ordered*)*. Then A*^{⊥} *is a prime ideal*.

*Proof*. Assume that *A* is an ideal which is linear, and *a* ∧ *b* ∈ *A*^{⊥} but *a, b* ∉ *A*^{⊥}. Then there are *x*′, *x*″ ∈ *A*, such that *a* ∧ *x*′ ≠ 0, and *b* ∧ *x*″ ≠ 0. Set *x* = *x*′∨ *x*″. Then *x* ∈ *A* as *A* is an ideal. Clearly, *a* ∧ *x* = *a* ∧ (*x*′ ∨ *x*″) = (*a* ∧ *x*^{′}) ∨(*a* ∧*x*″). Similarly, we have *b* ∧*x* ≠ 0. Since *a* ∧*x* ≤ *x, b* ∧*x* ≤ *x*, we conclude *a* ∧*x, b* ∧*x* ∈ *A*. As *A* is linear, we may assume that *a* ∧ *x* ≤ *b* ∧ *x*. Now, 0 = (*a* ∧ *b*) ∧ *x* = *a* ∧ (*b* ∧ *x*) ≥ *a* ∧ (*a* ∧ *x*) = *a* ∧ *x*, which contradicts the fact *a* ∧ *x* ≠ 0, which implies that *a* ∈ *A*^{⊥} or *b* ∈ *A*^{⊥}. Therefore, *A*^{⊥} is prime. □

*Let L be a BL-algebra, if X* ⊆ *Y* ⊆ *L, then Y*^{⊥} ⊆ *X*^{⊥};

*Proof*. If *z* ∈ *Y* ^{⊥}, we have *z* ∧ *y* = 0. Then for any *x* ∈ *X* ⊆ *Y*, *z* ∧ *x* = 0, and so *z* ∈∩_{x∈X} {*x*}^{⊥} = *X*^{⊥}. This means that *Y*^{⊥} ∨ *X*^{⊥}. □

*Let L be a BL-algebra, for any* ∅ ≠ *X* ⊆ *L, the following hold*:

*X*^{⊥} = ∩*x*∈*X* {*x*}^{⊥};

*X* ⊆ *X*^{⊥⊥};

*X*^{⊥} = *X*^{⊥⊥};

*X*^{⊥} = 〈*X*〉.^{⊥}

*Proof*. (1) *a* ∈ *X*^{⊥} ⇔ for all *x* ∈ *X*, *a* ∧ *x* = 0 ⇔ for all *x* ∈ *X*, *a* ∈ *x*^{⊥} ⇔ *a* ∈∩_{x∈X} {*x*}^{⊥}.

(2) By the definition of annihilator, we have *X*^{⊥⊥} = {a ∈ *L* | *a* ∧*x* = 0 *for all x* ∈ *X*^{⊥}}. So for all *x* ∈ *X*^{⊥}, if *b* ∈ *X*, then *b* ∧ *x* = 0, hence *b* ∈ *X*^{⊥⊥}.

(3) By (2) taking *X* = *X*^{⊥} we have *X*^{⊥} ⊆ *X*^{⊥⊥⊥}. Conversely, by (1) and Proposition 3.8, we have *X*^{⊥⊥⊥}⊆ *X*^{⊥}, therefore, *X*^{⊥} = *X*^{⊥⊥⊥}.

(4) Since *X* ⊆ 〈*X*〉, by Proposition 3.9, we have 〈*X*〉^{⊥}⊆ *X*^{⊥}. Now we prove *X*^{⊥}⊆ 〈*X*〉^{⊥}. Let *y* ∈ *X*^{⊥}, so for any *x* ∈ *X*, we have *x* ∧*y* = 0. For any *z* ∈ 〈*X*〉, there are *x*_{1},*x*_{2}, · · ·*x*_{n} ∈ *X*, such that *z* ≤ *x*_{1} ⊘*x*_{2} ⊘· · ·⊘ *x*_{n}. So *y* ∧ *z* ≤ *y* ∧ (*x*_{1} ⊘ *x*_{2} ⊘· · ·⊘ *x*_{n}) ≤ (*y* ∧ *x*_{1}) ⊘ (*y* ∧ *x*_{2}) ⊘· · ·⊘ (*y* ∧ *x*_{n}) = 0 ⊘ 0 ⊘· · ·⊘ 0 = 0. Hence *y* ∈ 〈*X*〉^{⊥}, that is, *X*^{⊥}⊆ 〈*X*〉^{⊥}. Therefore, *X*^{⊥} = 〈*X*〉^{⊥}. □

*Let L be a BL-algebra, and X, Y* ⊆ *L. Then*

*L*^{⊥} = {0};

*X*^{⊥⊥} *X*^{⊥} = {0};

(*X* ∪ *Y*)^{ ⊥} = *X*^{⊥} ∩ *Y*^{⊥};

*X*^{⊥} ∪ *Y* ^{⊥}⊆ (*X* ∩ *Y*)^{⊥}.

*Proof*. (1) If there is *a* ∈ *L* such that *a* ≠ 0 and *a* ∈ *L*^{⊥}, then 1 ∧ *a* = *a* ≠ 0, this is a contradiction, so *a* = 0.

(2) If *x* ∈ *X*^{⊥⊥} ∩ *X*^{⊥}, by the definition of annihilator, we have *x* = *x* ∧ *x* = 0, so *X*^{⊥⊥} ∩ *X*^{⊥} ⊆ 0. Conversely, since for every *Y* ⊆ *L* there is 0 ∈ *Y*^{⊥}, we get 0 ∈ *X*^{⊥⊥} and 0 ∈ *X*^{⊥}, so 0 ∈ *X*^{⊥⊥} ∩ *X*^{⊥}, finally *X*^{⊥⊥} ∩ *X*^{ ⊥} = {0}.

(3) Since *X* ⊆ *X* ∪ *Y* and *Y* ⊆ *X* ∪ *Y*, we have (*X* ∪ *Y*)^{ ⊥}⊆ *X*^{⊥} and (*X* ∪ *Y*)^{ ⊥} ⊆ *Y*^{⊥}, so (*X* ∪ *Y*)^{ ⊥}⊆ *X*^{⊥}∩ *Y*^{⊥}, conversely, for any *a* ∈ *X*^{⊥}∩ *Y* ^{⊥}, we have *a* ∈ *X*^{⊥} and *a* ∈ *Y*^{⊥}, ie. for any *x* ∈ *X, y* ∈ *Y*, we have *a* ∧ *x* = 0 and *a* ∧ *y* = 0. So for any *t* ∈ *X* ∪ *Y*, we always have *a* ∧ *t* = 0, hence *a* ∈ (*X* ∪ *Y*)^{ ⊥}.

(4) Since *X* ∩ *Y* ⊆ *X* and *X* ∩ *Y* ⊆ *Y*, we have *X*^{⊥} ⊆ (*X* ∩ *Y*)^{ ⊥} and *Y*^{⊥}⊆ (*X* ∩ *Y*)^{ ⊥}, so *X*^{⊥} ∪ *Y* ^{⊥} ⊆ (*X* ∩ *Y*)^{ ⊥}. □

*Let L be the BL-algebra in Example* 3.2*. Take X* ={*a, c*}*, Y* = {*c, b*}*, then X ∩ Y* = {*c*}*, so we have X*^{⊥} ={0}*, Y*^{⊥} = {0}*, and* (*X* ∩ *Y*)^{ ⊥} = {0*, a, b*}*. Hence* (*X* ∩ *Y*)^{ ⊥}⊈ *X*^{⊥} ∪ *Y*^{⊥} *. Therefore, we do not have the equation for Proposition* 3.11(4)

*Let L be a BL-algebra, and* ∅ *≠ X, Y* ⊆ *L. For all a, b* ∈ *L, if a* ∈ *X*^{⊥}, *b* ∈ *Y*^{⊥}*, then (1) a* ∧ *b* ∈ (*X* ∪ *Y*)^{ ⊥}*, (2) a* ∨ *b* ∈ (*X* ∩ *Y*) ^{⊥}.

*Proof*. (1) If *a* ∈ *X*^{⊥}, *b* ∈ *Y*^{⊥}, then *a* ∧ *b* ∈ *X*^{⊥} and *a* ∧ *b* ∈ *Y*^{⊥}, so *a* ∧ *b* ∈ *X*^{⊥} ∩ *Y*^{⊥} = (*X* ∪ *Y*)^{⊥}.

(2) If *a* ∈ *X*^{⊥}, *b* ∈ *Y* ^{⊥}, then *a* ∈ *X*^{⊥} ⊆ (*X* ∩ *Y*)^{⊥} and *b* ∈ *Y*^{⊥} ⊆ (*X* ∩ *Y*)^{ ⊥}, so *a* ∨ *b* ∈ (*X* ∩ *Y*)^{ ⊥}. □

*Let L be a BL-algebra, X, Y* ⊆ *L. Then X*^{⊥} ∩ *Y*^{⊥} ={0} *if and only if X*^{⊥} ⊆ *Y* ^{⊥⊥} *and Y*^{⊥} ⊆ *X*^{⊥⊥}.

*Proof*. ⇒ For all *a* ∈ *X*^{⊥}, *b* ∈ *Y* ^{⊥}, we have *a* ∧ *b* ∈ *X*^{⊥} ∩ *Y* ^{⊥}, since *X*^{⊥} ∩ *Y* ^{⊥} ={0}, we get *a* ∧ *b* = 0, by definition of annihilator, we get *a* ∈ *Y* ^{⊥⊥} and *b* ∈ *X*^{⊥⊥}, so *X*^{⊥} ⊆ *Y* ^{⊥⊥} and *Y* ^{⊥} ⊆ *X*^{⊥⊥}.

⇐ If *X*^{⊥} ⊆ *Y* ^{⊥⊥} and *Y* ^{⊥} ⊆ *X*^{⊥⊥}, then *X*^{⊥} ∩ *Y* ^{⊥} ⊆ *Y* ^{⊥⊥} ∩ *Y* ^{⊥} ={0}, so *X*^{⊥} ∩ *Y* ^{⊥} ⊆ {0}. Clearly, {0} ⊆ *X*^{⊥} ∩ *Y* ^{⊥}, so *X*^{⊥} ∩ *Y* ^{⊥} ={0}. □

*Let L be a BL-algebra, I be a non-empty subset of L. Then there is a X* ⊆ *L, satisfied I* = *X*^{⊥} *if and only if I is a down set, and I* ^{⊥⊥} = *I, I* ∩ *X*^{⊥⊥} ={0}*, X*^{⊥} ∩ *I* ^{⊥} ={0}.

*Proof*. ⇒If *I* = *X*^{⊥}, it is clear that *I* is a down set, *I* ^{⊥⊥} = *X*^{⊥⊥⊥} = *X*^{⊥} = *I*, *I* ∩ *X*^{⊥⊥} = *X*^{⊥} ∩ *X*^{⊥⊥} ={0}. Then we have *X*^{⊥} ∩ *I* ^{⊥} = *X*^{⊥} ∩ *X*^{⊥⊥} ={0} by Proposition 3.11 (2)

⇐ Since *I* ∩ *X*^{⊥⊥} ={0}, for any *i* ∈ *I*, for any *x* ∈ *X*^{⊥⊥}, then *i* ∧ *x* ∈ *I* ∩ *X*^{⊥⊥} ={0}, so *i* ∈ *X*^{⊥⊥⊥} . We have already known *X*^{⊥⊥⊥} = *X*^{⊥}, so *I* ⊆ *X*^{⊥}. Since *X*^{⊥} ∩ *I* ^{⊥} ={0}, then similarly, we get *X*^{⊥}⊆ *I* ^{⊥⊥}, since *I* ^{⊥⊥} = *I*, we get *X*^{⊥}⊆ *I*. Therefore, *I* = *X*^{⊥} . □

*Let L be a BL-algebra, if an ideal I which is linear contains an element x* ≠ 0 *and x* ∨ *x* = 1*, then x is the largest element of I*.

*Proof*. By *x* ∨ *x* = 1, we have *x* ∧ $\overline{\overline{x}}$ = 0. So *x* ∧ *x* ≤ *x* ∧ $\overline{\overline{x}}$ = 0. Let *a* ∈ *I*, then *a* = *a* ∧ 1 = *a* ∧ (*x* ∨ *x*) = (*a* ∧ *x*) ∨ (*a* ∧ *x*), where the last equation follows by the distributive of *L*. Since *I* is linear, by Proposition 3.8, we have *I* ^{⊥} is a prime ideal. Since *x* ∧ *x* = 0 ∈ *I* ^{⊥}, either *x* ∈ *I* ^{⊥} or *x* ∈ *I* ^{⊥}. As *x* ∧ *x* = *x* ≠ 0, we necessarily have *x* ∈ *I* ^{⊥}. Hence for any *a* ∈ *I*, we have *a* ∧ *x* = 0, which implies that *a* = *a* ∧ *x*, thus *a* ≤ *x*. Therefore, *x* is the largest element of *I*. □

*Let L be a BL-algebra. Then the ideal lattice I*(*L*) *is pseudo-complemented and for any ideal I of L, its pseudo-complement is I* ^{⊥}.

*Proof*. By Proposition 3.7, we have *I* ∩ *I* ^{⊥} ={0}. Let *G* be an ideal of *L* such that *I* ∩ *G* ={0}, we shall prove that *G* ⊆ *I* ^{⊥}. Let *a* ∈ *G*, for any *x* ∈ *I*, then we have *x* ∧ *a* ≤ *x* ∈ *I*, *x* ∧ *a* ≤ *a* ∈ *G*, so *x* ∧ *a* ∈ *I* ∩ *G* ={0}. Hence *x* ∧ *a* = 0 for any *x* ∈ *I*, then we have *a* ∈ *I* ^{⊥}. So *I* ^{⊥} is the largest ideal such that *I* ∩ *G* ={0}. It follows that *I* ^{⊥} is the pseudo-complement of *I*. □

Let *An*(*L*) ={*X*^{⊥} | *X* ⊆ *L*} be the set of annihilators of *L*. Since *X*^{⊥} = 〈*X*〉^{⊥}, we get that *An*(*L*) = {*I* ^{⊥} | *I* ∈ *I*(*L*)}. Hence *An*(*L*) is the set of pseudo-complements of the pseudo-complemented lattice *I*(*L*).

*Let L be a BL-algebra, I, J* ∈ *I*(*L*)*. Then*

{0}*;L* ∈ *An*(*L*);

*I* ∈ *An*(*L*) ⇔ *I*^{⊥⊥} = *I ;*

⊥⊥: *X* → *X*^{⊥⊥} *is a closure map;*

*I* ∩ (*I* ∩ *J*)^{⊥} = *I* ∩ *J* ^{⊥};

(*I* ∩ *J*)^{⊥⊥} = *I* ^{⊥⊥} ∩ *J* ^{⊥⊥};

*I, J* ∈ *An*(*L*)*, then I* ∧ _{An(L)} *J* = *I* ∩ *J ;*

(*I* ∨ *J*)^{⊥} = *I* ^{⊥} ∩ *J* ^{⊥};

*if I, J* ∈ *An*(*L*)*, then I* ∨_{An(L)} *J* = (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥}.

*Proof*. (1) By Propositions 3.4 *and* 3.11.

(2) Assume that *I* ∈ *An*(*L*), then there exists *X* ⊆ *L* such that *X*^{⊥} = *I*, so we get *I* ^{⊥⊥} = *X*^{⊥⊥⊥} = *X*^{⊥} = *I*. The converse is clear.

(3) By Propositions 3.9, we know the function *f* : *X* → *X*^{⊥⊥} is isotone and by Propositions 3.10, we get that *f* = *f* ^{2} ≥ *id*_{L}. So, *X* → *X*^{⊥⊥} is a closure map.

(4) Since (*I* ∩ *J*) ∩ (*I* ∩ *J*)^{⊥} ={0}, by Proposition 3.17, we get *I* ∩ (*I* ∩ *J*)^{⊥} ∨ *J* ^{⊥} and so *I* ∩ (*I* ∩ *J*)^{⊥} ⊆ *I* ∩ *J* ^{⊥}. Conversely, by *I* ∩ *J* ⊆ *J*, we get *J* ^{⊥} ⊆ (*I* ∩ *J*)^{⊥}, so *I* ∩ *J* ^{⊥} ∨ *I* ∩ (*I* ∩ *J*)^{⊥}. Therefore, *I* ∩ (*I* ∩ *J*)^{⊥} = *I* ∩ *J* ^{⊥}.

(5) Since *I* ∩ *J* ⊆ *I, J*, we get (*I* ∩ *J*)^{⊥⊥} ⊆ *I* ^{⊥⊥} ∩ *J* ^{⊥⊥}. Conversely, (*I* ∩ *J*) ∩ (*I* ∩ *J*)^{⊥} ={0}) ⇒ *I* ∩ (*I* ∩ *J*)^{⊥} ⊆ *J* ^{⊥} = *J* ^{⊥⊥⊥} ⇒ *I* ∩ *J* ^{⊥⊥} ∩ (*I* ∩ *J*)^{⊥} ={0} ⇒ *J* ^{⊥⊥} ∩ (*I* ∩ *J*)^{⊥} ⊆ *I* ^{⊥} = *I* ^{⊥⊥⊥})*I* ^{⊥⊥} ∩ *J* ^{⊥⊥} ∩ (*I* ∩ *J*)^{⊥} ={0}) *I* ^{⊥⊥} ∩ *J* ^{⊥⊥}⊆ (*I* ∩ *J*)^{⊥⊥}. So we get (*I* ∩ *J*)^{⊥⊥} = *I* ^{⊥⊥} ∩ *J* ^{⊥⊥}.

(6) By (3) and Proposition 2.6, we have *I* ∧ _{An(L)} *J* = *I* ∩ *J*.

(7) Since *I, J* ⊆ *I* ∨ *J*, we get (*I* ∨ *J*)^{⊥} ⊆ *I* ^{⊥} ∩ *J* ^{⊥} = *I* ^{⊥⊥⊥} ∩ *J* ^{⊥⊥⊥} = (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥⊥}. Conversely, *I* ∨ *I* ^{⊥⊥}⊆ (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥}, similarly, we have *J* ⊆ *J* ^{⊥⊥}⊆ (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥}, so *I* ∨ *J* ⊆ (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥}, hence (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥⊥}⊆ (*I* ∨ *J*)^{⊥}. Therefore, (*I* ∨ *J*)^{⊥} = (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥⊥} = *I* ^{⊥⊥⊥} ∩ *J* ^{⊥⊥⊥} = *I* ^{⊥} ∩ *J* ^{⊥}.

(8) By (3) and Proposition 2.6, we have *I* ∨_{An(L)} *J* = (*I* ∨ *J*)^{⊥⊥}, then by (7) we have *I* ∨_{An(L)} *J* = (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥} . □

*Let L be a BL-algebra. Then* (*An*(*L*); ∩ ; ∨_{An(L)}, ⊥, {0}*; L*) *is a Boolean algebra*.

*Proof*. Firstly, we show that *An*(*L*) is distributive, it suffices to prove that: for all *I, J, H* ∈ *An*(*L*), *H* ∩ (*I* ∨_{An(L)} *J*) ⊆ (*H* ∩ *I*) ∨_{An(L)} (*H* ∩ *J*). Now, let *K* = (*H* ∩ *I*) ∨_{An(L)} (*H* ∩ *J*), then *H* ∩ *I* ⊆ *K* = *K*^{⊥⊥} gives *H* ∩ *I* ∩ *K*^{⊥} ={0} and so *H* ∩ *K*^{⊥}⊆ *I* ^{⊥}. Similarly, *H* ∩ *K*^{⊥}⊆ *J* ^{⊥} and therefore *H* ∩ *K*^{⊥}⊆ *I* ^{⊥} ∩ *J* ^{⊥} = (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥⊥}. It follows that *H* ∩ *K*^{⊥} ∩ (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥} ={0} and hence *H* ∩ (*I* ∨_{An(L)} *J*) = *H* ∩ (*I* ^{⊥} ∩ *J* ^{⊥})^{⊥} ⊆ *K*^{⊥⊥} = *K* = (*H* ∩ *I*) ∨_{An(L)} (*H* ∩ *J*).

Secondly, we show that *An*(*L*) is complemented, observe that *L* ={0}^{⊥} ∈ *An*(*L*) and {0}= *L*^{⊥} ∈ *An*(*L*). Since for every *I* ∈ *An*(*L*) we have *I* ∩ *I* ^{⊥} ={0} and *I* ∨_{An(L)} *I* ^{⊥} = (*I* ^{⊥} ∩ *I* ^{⊥⊥})^{⊥} ={0}^{⊥} = *L*. So the complement of *I* ∈ *An*(*L*) is *I* ^{⊥}. Therefore, *An*(*L*) is a Boolean algebra. □

*Let L be a BL-algebra*, ∅ ≠ *X* ⊆ *L, I be an ideal of L and f be an endomorphism. We define the annihilator of X with respect to I to be the set* ${X}_{I}^{{\perp}_{f}}$ ={*a* ∈ *L* | *f* (*a*) ∧ *x* ∈ *I*, ∀*x* ∈ *X*}.

If *f* = *id*_{L}, we denote ${X}_{I}^{\perp}:={X}_{I}^{\perp i{d}_{L}}$ ={*a* ∈ *L* | *a* ∧ *x* ∈ *I*, ∀*x* ∈ *X*}.

*Let L* ={0*, a, b, c*, 1} *be a set, where* 0 ≤ *a* ≤ *c* ≤ 1 *and* 0 ≤ *b* ≤ *c* ≤ 1*. The Cayley tables are as follows*:

*Then* (*L*, ∧, ∨, ⊙, →, 0, 1) *is a BL-algebra. Now we define a map f as follows: f* (0) = 0*, f* (*a*) = *b, f* (*b*) = *a, f* (*c*) = *c, f* (1) = 1*, then we can check that f is an endomorphism. Let X* ={*a, c*}*, and I* ={0*, a*}*, then we get* ${X}_{I}^{{\perp}_{f}}$ = {0, *b*}.

*Let L* = {0*, a, b*, 1} *be a set, where* 0 ≤ *a, b* ≤ 1*. The Cayley tables are as follows*:

*Then* (*L*, ∧, ∨, ⊙, →, 0, 1) *is a BL-algebra. Let X* = {0, *b*}*, and I* = {0*, a*}*, then we get* ${X}_{I}^{\perp}$ = {0*, a*}.

*Let L be a BL-algebra*, ∅ ≠ *X* ⊆ *L, I be an ideal of L and f be an endomorphism. Then* ${X}_{I}^{{\perp}_{f}}$ *is an ideal of L*.

*Proof*. Clearly, 0 ∈ ${X}_{I}^{{\perp}_{f}}$, so ${X}_{I}^{{\perp}_{f}}$ is nonempty.

I1: Let *a* ∈ ${X}_{I}^{{\perp}_{f}}$ and *b* ∈ *L* such that *b* ≤ *a*. Then *f* (*b*) ≤ *f* (*a*). It follows that *f* (*b*) ∧ *x* ≤ *f* (*a*) ∧ *x*. Since *f* (*a*) ∧ *x* ∈ *I*, we have *f* (*b*) ∧ *x* ∈ *I*, which implies *b* ∈ ${X}_{I}^{{\perp}_{f}}$;

I2: Let *a, b* ∈ ${X}_{I}^{{\perp}_{f}}$, then *f* (*a*)∧*x* ∈ *I* and *f* (*b*)∧*x* ∈ *I* for all *x* ∈ *X*. Since *f* (*a*⊘*b*)∧*x* = (*f* (*a*)⊘*f* (*b*))∧*x* ≤ (*f* (*a*) ∧ *x*) ⊘ (*f* (*b*) ∧ *x*), we have *f* (*a* ⊘ *b*) ∧ *x* ∈ *I*, that is, *a* ⊘ *b* ∈ ${X}_{I}^{{\perp}_{f}}$. □

*Let L be a BL-algebra, f be an endomorphism, I, J* ∈ *I*(*L*) *and* ∅ ≠ *X, X*′ ⊆ *L. Then we have*:

*I* ⊆ *J implies* ${X}_{I}^{{\perp}_{f}}\subseteq {X}_{J}^{{\perp}_{f}};$

*X* ⊆ *X*′ *implies* ${\left({X}^{\prime}\right)}_{I}^{{\perp}_{f}}\subseteq {X}_{I}^{{\perp}_{f}};$

${\left({\cup}_{\lambda \in \Lambda}{X}_{\lambda}\right)}_{I}^{{\perp}_{f}}={\cap}_{\lambda \in \Lambda}{\left({X}_{\lambda}\right)}_{I}^{{\perp}_{f}};$

${X}_{I}^{{\perp}_{f}}={\cap}_{x\in X}{x}_{I}^{{\perp}_{f}};$

${X}_{{\cap}_{\lambda \in \Lambda}{I}_{\lambda}}^{{\perp}_{f}}={\cap}_{\lambda \in \Lambda}{X}_{I}^{{\perp}_{f}};$

${\u3008X\u3009}_{I}^{{\perp}_{f}}={X}_{I}^{{\perp}_{f}};$

$Ker\left(f\right)\subseteq {X}_{\left\{0\right\}}^{{\perp}_{f}},{L}_{\left\{0\right\}}^{{\perp}_{f}}=Ker\left(f\right).$

*Proof*. (1) Let *I* ⊆ *J* and *a* ∈ ${X}_{I}^{{\perp}_{f}}$. Then *f* (*a*) ∧ *x* ∈ *I* for all *x* ∈ *X*. So we have *f* (*a*) ∧ *x* ∈ *J* for all *x* ∈ *X*, that is, *a* ∈ ${X}_{J}^{{\perp}_{f}}$. Therefore, ${X}_{I}^{{\perp}_{f}}$ ⊆ ${X}_{J}^{{\perp}_{f}}$.

(2) Let *X* ⊆ *X*′ and *a* ∈ ${({X}^{\prime})}_{I}^{{\perp}_{f}}$. Then *f* (*a*) ∧*x* ∈ *I* for all *x* ∈ *X*. Hence *f* (*a*) ∧*x* ∈ *I* for all *x* ∈ *X*, which implies *a* ∈ ${X}_{I}^{{\perp}_{f}}$. Therefore, ${({X}^{\prime})}_{I}^{{\perp}_{f}}$ ⊆ ${X}_{I}^{{\perp}_{f}}$.

(3) By (2) we have ${\left({\cup}_{\lambda \in \Lambda}{X}_{\lambda}\right)}_{I}^{{\perp}_{f}}\subseteq {\left({X}_{\lambda}\right)}_{I}^{{\perp}_{f}}$ for all *λ* ∈⋀, so we get ${\left({\cup}_{\lambda \in \Lambda}{X}_{\lambda}\right)}_{I}^{{\perp}_{f}}\subseteq {\cap}_{\lambda \in \Lambda}{\left({X}_{\lambda}\right)}_{I}^{{\perp}_{f}}$. Conversely, let $a\in {\cap}_{\lambda \in \Lambda}{\left({X}_{\lambda}\right)}_{I}^{{\perp}_{f}}$, we have $a\in {\left({X}_{\lambda}\right)}_{I}^{{\perp}_{f}}$ for all *λ* ∈Λ. Hence *f* (*a*) ∧ *x*_{λ} ∈ *I* for all *x*_{λ} ∈ *X*_{λ} and

*λ* ∈ Λ, which implies $a\in {\left({\cup}_{\lambda \in \Lambda}{X}_{\lambda}\right)}_{I}^{{\perp}_{f}}$. Therefore, ${\left({\cup}_{\lambda \in \Lambda}{X}_{\lambda}\right)}_{I}^{{\perp}_{f}}={\cap}_{\lambda \in \Lambda}{\left({X}_{\lambda}\right)}_{I}^{{\perp}_{f}}$.

(4) It is clear by (3).

(5) We have *a* ∈ ${X}_{{\cap}_{\lambda \in \Lambda}{I}_{\lambda}}^{{\perp}_{f}}$ if and only if *f* (*a*) ∧ *x* ∈ ∩ _{λ∈}Λ *I*_{λ} for all *x* ∈ *X*, and if and only if *f* (*a*) ∧ *x* ∈ *I*_{λ} for all *x* ∈ *X* and *λ* ∈ Λ, which is equivalent to *a* ∈ ${X}_{I}^{{\perp}_{f}}$ for all *λ* ∈ Λ, that is, *a* ∈ ${\cap}_{\lambda \in \Lambda}{X}_{{I}_{\lambda}}^{{\perp}_{f}}$.

(6) Since *X* ⊆ 〈*X*〉, by (2) we get ${\u3008X\u3009}_{I}^{{\perp}_{f}}\subseteq {X}_{I}^{{\perp}_{f}}$. Conversely, let *a* ∈ ${X}_{I}^{{\perp}_{f}}$ and *z* ∈ 〈*X*〉. Then *f* (*a*) ∧ *x* ∈ *I* for all *x* ∈ *X*. Since *z* ∈ 〈*X*〉, then there exist *x*_{1}*;x*_{2}; ··· *;x*_{n} ∈ *X* such that *z* ≤ *x*_{1}, ⊘*x*_{2} ⊘···⊘ *x*_{n}. It follows that *f* (*a*) ∧ *z* ≤ *f* (*a*) ∧ (*x*_{1} ⊘ *x*_{2} ⊘···⊘ *x*_{n}) ≤ (*f* (*a*) ∧ *x*_{1}) ⊘ (*f* (*a*) ∧ *x*_{2}) ⊘···⊘(*f* (*a*) ∧ *x*_{n}), since *f* (*a*) ∧ *x*_{i} ∈ *I* for all 1 ≤ *i* ≤ *n*, we get *f* (*a*) ∧ *z* ∈ *I*, which implies that *a* ∈ 〈*X*〉${}_{I}^{{\perp}_{f}}$. Therefore, 〈*X*〉${}_{I}^{{\perp}_{f}}$ = *X*${}_{I}^{{\perp}_{f}}$.

(7) Let *a* ∈ Ker(f), we have *f* (*a*) = 0, then *f* (*a*) ∧ *x* = 0 for all *x* ∈ *X*, that is, *a* ∈ *X*${}_{\left\{0\right\}}^{{\perp}_{f}}$, hence Ker(f) ⊆ *X*${}_{\left\{0\right\}}^{{\perp}_{f}}$. Let *a* ∈ *L*${}_{\left\{0\right\}}^{{\perp}_{f}}$, then *f* (*a*) ∧ *x* = 0 for all *x* ∈ *L*. In particular, taking *x* = *f* (*a*), we have *f* (*a*) = 0, which implies *a* ∈ Ker(f). Hence *L* ${}_{\left\{0\right\}}^{{\perp}_{f}}$ ⊆ Ker(f). Conversely, by Ker(f) ⊆ *X*${}_{\left\{0\right\}}^{{\perp}_{f}}$, for any ∅ ≠ *X* ⊆ *L*, taking *X* = *L*, so Ker(f) ⊆ *L*${}_{\left\{0\right\}}^{{\perp}_{f}}$. Therefore, *L*${}_{\left\{0\right\}}^{{\perp}_{f}}$ =Ker(f). □

*Let I be an ideal of a BL-algebra L, f be an endomorphism and a, b* ∈ *L. Then* (1) *a* ≤ *b implies b*${}_{I}^{{\perp}_{f}}$ ⊆ *a*${}_{I}^{{\perp}_{f}}$ *; (2)* (*a* ∨ *b*)${}_{I}^{{\perp}_{f}}$ = *a*${}_{I}^{{\perp}_{f}}$ ∩ *b*${}_{I}^{{\perp}_{f}}$.

*Proof*. (1) Let *x* ∈ *b*${}_{I}^{{\perp}_{f}}$, then *f* (*x*) ∧ *b* ∈ *I* and *f* (*x*) ∧ *a* ≤ *f* (*x*) ∧ *b*, since *a* ≤ *b*. It follows that *f* (*x*) ∧ *a* ∈ *I*, that is, *x* ∈ *a* ${}_{I}^{{\perp}_{f}}$;

(2) Since *a, b* ≤ *a* ∨ *b*, by (1) we have that (*a* ∨ *b*)${}_{I}^{{\perp}_{f}}$ ⊆ *a*${}_{I}^{{\perp}_{f}}$, *b*${}_{I}^{{\perp}_{f}}$, so (*a* ∨ *b*)${}_{I}^{{\perp}_{f}}$ ⊆ *a*${}_{I}^{{\perp}_{f}}$ ∩ *b*${}_{I}^{{\perp}_{f}}$. Conversely, let *x* ∈ *a* ${}_{I}^{{\perp}_{f}}$ ∩ *b*${}_{I}^{{\perp}_{f}}$, that is, *a* ∧ *f* (*x*), *b* ∧ *f* (*x*) ∈ *I*, it follows *f* (*x*) ∧ (*a* ∨*b*) = (*f* (*x*) ∧ *a*) ∨ (*f* (*x*) ∧ *b*) ∈ *I*, as *I* is an ideal of *L*, and *L* is a distributive lattice. Therefore, *x* ∈ (*a* ∨ *b*)${}_{I}^{{\perp}_{f}}$; □

*Let L be a BL-algebra*, ∅ ≠ *X* ⊆ *L, I be an ideal of L. Then*

*I* ⊆ ${X}_{I}^{\perp}$;

${X}_{I}^{\perp}$ = *L if and only if X* ⊆ *I*.

*Proof*. (1) Let *i* ∈ *I*, then *i* ∧ *x* ≤ *i* ∈ *I* for all *x* ∈ *X*, hence *i* ∈ ${X}_{I}^{\perp}$, that is, *I* ⊆ ${X}_{I}^{\perp}$;

(2) If ${X}_{I}^{\perp}$ = *L*, then 1 ∈ ${X}_{I}^{\perp}$, so for all *x* ∈ *X* we have *x* = *x* ∧ 1 ∈ *I*. Conversely, if *X* ⊆ *I*, then for any *a* ∈ *L* and for all *x* ∈ *X* we have *a* ∧ *x* ≤ *x* ∈ *I*, so *a* ∈ ${X}_{I}^{\perp}$. That is, *L* ⊆ ${X}_{I}^{\perp}$, which implies that ${X}_{I}^{\perp}$ = *L*. □

*Let L be a BL-algebra, I, J, H* ∈ *I*(*L*)*. Then we have*:

(1) ${J}_{I}^{\perp}$ ∩ *J* ⊆ *I ;*

(2) *J* ∩ *H* ⊆ *I if and only if H* ⊆ ${J}_{I}^{\perp}$.

*Proof*. (1) Let *x* ∈ ${J}_{I}^{\perp}$ ∩ *J*, then *x* ∈ ${J}_{I}^{\perp}$ and *x* ∈ *J*, so we get *x* = *x* ∧ *x* ∈ *I*. That is, ${J}_{I}^{\perp}$ ∩ *J* ⊆ *I* ;

(2) If *J* ∩ *H* ⊆ *I* and *x* ∈ *H*, then for any *y* ∈ *J*, we have *x* ∧ *y* ∈ *J* ∩ *H*, it follows that *x* ∧ *y* ∈ *I*. Hence *x* ∈ ${J}_{I}^{\perp}$, that is *H* ⊆ ${J}_{I}^{\perp}$. Conversely, let *H* ⊆ ${J}_{I}^{\perp}$, by (1), we have *J* ∩ *H* ⊆ *J* ∩ ${J}_{I}^{\perp}$ ⊆ *I*. □

*Let L be a BL-algebra, I, J* ∈ *I*(*L*)*. Then* ${J}_{I}^{\perp}$ *is the relative pseudo-complement of J with respect to I in the lattice* (*I*(*L*), ⊆).

*Proof*. We have already known ${J}_{I}^{\perp}$ is an ideal, and by Proposition 3.27 (1)*, we have* ${J}_{I}^{\perp}$ ∩ *J* ⊆ *I*. Now we show that ${J}_{I}^{\perp}$ is the greatest ideal of L such that *H* ∩ *J* ⊆ *I*, where *H* ∈ *I*(*L*). Assume that *H* is an ideal of *L* such that *H* ∩ *J* ⊆ *I*, If *a* ∈ *H*, then *a* ∧ *x* ≤ *a, x* for all *x* ∈ *J*. Since *J* and *H* are ideals of *L*, we have *a* ∧ *x* ∈ *J* ∩ *H*, Hence *a* ∧ *x* ∈ *I* for all *x* ∈ *J*, that is, *a* ∈ ${J}_{I}^{\perp}$. Therefore, ${J}_{I}^{\perp}$ is the relative pseudo-complement of *J* with respect to *I* in the lattice (*I*(*L*), ⊆). □

*Let L, M be two BL-algebras, f* : *L* → *M be a homomorphism*, ∅ ≠ *A* ⊆ *L. Then f* (*A*^{⊥}) ⊆ (*f* (*A*))^{⊥}.

*Proof*. For all *x* ∈ *f* (*A*^{⊥}), there is a *y* ∈ *A*^{⊥}, such that *x* = *f* (*y*). For all *z* ∈ *f* (*A*), there is a *t* ∈ *A*, such that *z* = *f* (*t*). So we have *x* ∧ *z* = *f* (*y*) ∧ *f* (*t*) = *f* (*y* ∧ *t*) = *f* (0) = 0, therefore, *x* ∈ (*f* (*A*))^{⊥} . □

The next example shows the following: let *L*, *M* be two BL-algebras, *f* : *L* → *M* be a homomorphism, ∅ ≠ *A* ⊆ *L*, then *f* (*A*^{⊥}) may not be an annihilator of a subset of *M*.

*Let L be the BL-algebra of Example* 3.22*. Let* $M=\{0,\frac{1}{2},1\}$*, such that* $0\le \frac{1}{2}\le 1$*. The Cayley tables are as follows*:

*Then* (*M*, ∧, ∨, ⊙, →, 0, 1) *is a BL-algebra. Let f* (1) = *f* (*a*) = 1*, f* (0) = *f* (*b*) = 0*, then f* : *L* → *M is a homomorphism. Let A* ={b}*, then A*^{⊥} = {0, *a*}*, f* (*A*^{⊥}) = {0, 1}*, clearly* {0, 1} *is not a down set, so there is no B* ⊆ *M, such that f* (*A*^{⊥}) = *B*^{⊥}.

*Let L, M be two BL-algebras, f* : *L* → *M be a surjective homomorphism*, ∅ ≠ *B* ⊆ *M. Then* (*f* ^{1}(*B*))^{⊥} ⊆ *f* ^{1}(*B*^{⊥}).

*Proof*. For all *x* ∈ (*f* ^{1}(*B*))^{⊥}, and for all *b* ∈ *B*, there is a *a* ∈ *L*, such that *b* = *f* (*a*), so *x* ∧ *a* = 0, then *f* (*x*) ∧ *b* = *f* (*x*) ∧ *f* (*a*) = *f* (*x* ∧ *a*) = *f* (0) = 0. Therefore, *f* (*x*) ∈ *B*^{⊥}, which implies that *x* ∈ *f* ^{1}(*B*^{⊥}). □

The next example shows the following: let *L*, *M* be two BL-algebras, *f* : *L* → *M* be a surjective homomorphism, ∅ ≠ *B* ⊆ *M*, then *f* ^{1}(*B*^{⊥}) may not be an annihilator of a subset of *L*.

*Let L* = {0*, a, b, c, d*, 1}*, where* 0 ≤ *a* ≤ *c* ≤ 1, 0 ≤ *b* ≤ *d* ≤ 1 *and* 0 ≤ *b* ≤ *c* ≤ 1*. The Cayley tables are as follows*:

*Then* (*L*, ∧, ∨, ⊙, →, 0, 1) *is a BL-algebra*.

*Let M* = {0, 1}*. The Cayley tables are as follows*:

*Then* (*M*, ∧, ∨, ⊙, →, 0, 1) *is a BL-algebra. Let f* (0) = *f* (*b*) = *f* (*d* ) = 0*, f* (*a*) = *f* (*c*) = *f* (1) = 1*, then f* : *L* → *M is a homomorphism. Let B* ={1}*. Then B*^{⊥} ={0}*, f* ^{1}(*B*^{⊥}) = *f* ^{1}(0) = {0*; b; d* }*, we can check that there is no A* ⊆ *L, such that f* ^{1}(*B*^{⊥}) = *A*^{⊥}.

*Let L, M be two BL-algebras, f* : *L* → *M be a homomorphism*, ∅ ≠ *A* ⊆ *L. Then f* (*A*^{⊥}) = (*f* (*A*))^{⊥} *if and only if* (*f* (*A*^{⊥}))^{⊥⊥} = *f* (*A*^{⊥}) *and* (*f* (*A*))^{⊥} ∩ (*f* (*A*^{⊥}))^{⊥} ={0}.

*Proof*. ⇒ If *f* (*A*^{⊥}) = (*f* (*A*))^{⊥}, then (*f* (*A*^{⊥}))^{⊥⊥} = (*f* (*A*))^{⊥⊥⊥} = (*f* (*A*))^{⊥} = *f* (*A*^{⊥}), (*f* (*A*))^{⊥} ∩ (*f* (*A*^{⊥}))^{⊥} = (*f* (*A*))^{⊥} ∩ (*f* (*A*))^{⊥⊥} ={0}.

⇐ By Proposition 3.27, we have *f* (*A*^{⊥}) ⊆ (*f* (*A*))^{⊥} . Now we prove that *f* (*A*^{⊥}) *∩* (*f* (*A*))^{⊥} . Since (*f* (*A*))^{⊥} ∩ (*f* (*A*^{⊥}))^{⊥} ={0}, we have (*f* (*A*))^{⊥} ⊆ (*f* (*A*^{⊥}))^{⊥⊥} = *f* (*A*^{⊥}). Therefore, *f* (*A*^{⊥}) = (*f* (*A*))^{⊥} . □

*Let L, M be two BL-algebras, f* : *L* → *M be a surjective homomorphism*, ∅ ≠ *B* ⊆ *M. Then* (*f* ^{1}(*B*))^{⊥} = *f* ^{1}(*B*^{⊥}) *if and only if f* ^{1}(*B*^{⊥}) ∩ (*f* ^{1}(*B*))^{⊥⊥} ={0}.

*Proof*. ⇒ If (*f* ^{1}(*B*))^{⊥} = *f* ^{1}(*B*^{⊥}), then *f* ^{1}(*B*^{⊥}) ∩ (*f* ^{1}(*B*))^{⊥⊥} = (*f* ^{1}(*B*))^{⊥} ∩ (*f* ^{1}(*B*))^{⊥⊥} ={0}.

⇐ Firstly, if *x* ∈ *f* ^{1}(*B*^{⊥}), *y* ∈ *L*, such that *y* ≤ *x*, then *f* (*y*) ≤ *f* (*x*), since *f* (*x*) ∈ *B*^{⊥}, we get *f* (*y*) ∈ *B*^{⊥}, so *y* ∈ *f* ^{1}(*B*^{⊥}), and so we have *f* ^{1}(*B*^{⊥}) is a down set. Since *f* ^{1}(*B*^{⊥}) ∩ (*f* ^{1}(*B*))^{⊥⊥} ={0}, we get *f* ^{1}(*B*^{⊥}) ⊆ (*f* ^{1}(*B*))^{⊥⊥⊥} = (*f* ^{1}(*B*))^{⊥}, by Proposition 3.29, we already have (*f* ^{1}(*B*))^{⊥} ⊆ *f* ^{1}(*B*^{⊥}). Therefore, (*f* ^{1}(*B*))^{⊥} = *f* ^{1}(*B*^{⊥}). □

*Let L, M be two BL-algebras, f* : *L* → *M be an isomorphism*, ∅ ≠ *A* ⊆ *L*, ∅ ≠ *B* ⊆ *M. Then f* (*A*^{⊥}) = (*f* (*A*))^{⊥} *and* (*f* ^{1}(*B*))^{⊥} = *f* ^{1}(*B*^{⊥}).

*Proof*. (1) By Proposition 3.29, we have already known *f* (*A*^{⊥}) ⊆ (*f* (*A*))^{⊥}, we now prove *f* (*A*^{⊥}) ⊇ (*f* (*A*))^{⊥} . For any *y* ∈ (*f* (*A*))^{⊥}, since *f* is surjective, there is a *x* ∈ *L*, such that *f* (*x*) = *y*. For any *a* ∈ *A*, we have *f* (*a*) ∈ *f* (*A*), so *f* (*x* ∧ *a*) = *f* (*x*) ∧ *f* (*a*) = *y* ∧ *f* (*a*) = 0. Since *f* is injective, we get *x* ∧ *a* = 0, so *x* ∈ *A*^{⊥}, that is *y* ∈ *f* (*A*^{⊥}). Therefore, *f* (*A*^{⊥}) ⊇ (*f* (*A*))^{⊥}.

(2) By Proposition 3.31, we have already known (*f* ^{1}(*B*))^{⊥} ∨ *f* ^{1}(*B*^{⊥}), we now prove (*f* ^{1}(*B*))^{⊥} ⊇ *f* ^{1}(*B*^{⊥}). For any *x* ∈ *f* ^{1}(*B*^{⊥}), then *f* (*x*) ∈ *B*^{⊥}. For all *a* ∈ *f* ^{1}(*B*), so *f* (*a*) ∈ *B*, *f* (*x* ∧ *a*) = *f* (*x*)∧*f* (*a*) = 0, since *f* is injective, we get *x* ∧ *a* = 0, so *x* ∈ (*f* ^{1}(*B*))^{⊥}. Therefore, *f* ^{1}(*B*^{⊥}) ∨ (*f* ^{1}(*B*))^{⊥} . □

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