Let *P = {a*_{1},..., a_{n}} and *Q = {b*_{1},..., b_{m}} be two finite partitions of a quantum logic corresponding to a state s. The common refinement of these partitions is:

*P ∨ Q = {a*_{i} ∧ b_{j} : a_{i} ε P, b_{j} ε R}.

**Definition 7**. Let *P = {a*_{1},..., a_{n}} be a partition of a quantum logic corresponding to a state s. The logical entropy of *P* with respect to state s is defined by:
$${h}_{s}^{l}\left(p\right)={\displaystyle \sum _{i=1}^{n}s\left({a}_{i}\right)\left(1-s\left({a}_{i}\right)\right)}.$$
Since
${\sum}_{i=1}^{n}s({a}_{i})=1,$ we have
$\sum}_{i=1}^{n}{(s({a}_{i}))}^{2$

**Definition 8**. Let *P = {a*_{1},..., a_{n}} and *Q* ={b_{1},..., b_{m}} be two partitions of a quantum logic corresponding to a state *s*. The conditional logical entropy of *P* given *Q* with respect to state *s* is defined as:
$$$$
In the next theorem an upper bound for logical entropy on a quantum logic is presented.

**Theorem 9.** *Let P be a finite partition of a quantum logic corresponding to a state s. Then*
$$0\le 0{h}_{s}^{\mathrm{l}}(P)\le 1-\frac{1}{n}.$$

*Proof.* Let *P = (p*_{1},...,p_{n}) ε **R**^{n} be a probability distribution, then from [7], maximum value of the logical entropy is $0\le {h}_{s}^{l}(P)\le 1-\frac{1}{n}$
$$1-\frac{1}{n}.{\displaystyle {\sum}_{i=1}^{n}s({a}_{i})=1,}P=(s({a}_{1}),\mathrm{.....},s({a}_{n}))$$
is a probability distribution and hence the proof is complete.

In the following theorem the conditional logical entropy under the common refinement of partitions is studied.

**Theorem 10.** *Let P, Q and R be finite partitions of a quantum logic corresponding to a state s having Bayes’ Property. Then h*^{l}_{s}(P ∨ Q|R) = *h*^{1}_{s}(P|R) + *h*^{1}_{s}(Q|P ∨ R).

*Proof.* Let *P = {a*_{1},..., a_{n}}, Q = {b_{1},..., b_{m}} and *R = {c*_{1},..., c_{r}}. Since s has Bayes’ Property, by Lemma 6 we have
$$\begin{array}{l}{h}_{s}^{l}(P\vee Q|R)={\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}s({a}_{i}\wedge {b}_{j}\wedge {c}_{k})(s({c}_{k})-s({a}_{i}\wedge {b}_{j}\wedge {c}_{k}))\\ ={\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}s({a}_{i}\wedge {b}_{j}\wedge {c}_{k})(s({c}_{k})\\ -{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}(s({a}_{i}\wedge {b}_{j}\wedge {c}_{k}){(}^{s}\\ ={\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}s({a}_{i}\wedge {b}_{j}\wedge {c}_{k})(s({c}_{k})-{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}(s({a}_{i}\wedge {b}_{j}\wedge {c}_{k}){(}^{s}\\ +{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{k=1}^{r}}{(}^{s}\\ -{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}(s({a}_{i}\wedge {b}_{j}\wedge {c}_{k}){(}^{s}.\end{array}$$
On the other hand we can write
$${h}_{s}^{l}(P|R)={\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{k=1}^{r}}s({a}_{i}\wedge {b}_{j}\wedge {c}_{k})(s({c}_{k})-{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{k=1}^{r}}{(}^{s}$$
and
$$\begin{array}{l}{h}_{s}^{l}(Q|P\vee R)={\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}s({a}_{i}\wedge {b}_{j}\wedge {c}_{k})(s({c}_{k})(s({a}_{i}\wedge {c}_{k})-{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}(s({a}_{i}\wedge {b}_{j}\wedge {c}_{k}){(}^{s}.\\ ={\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{k=1}^{r}}{(}^{s}\\ -{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{\displaystyle {\sum}_{k=1}^{r}}(s({a}_{i}\wedge {b}_{j}\wedge {c}_{k}){(}^{s}\end{array}$$
Thus the proof is complete. □

Now the assertion of the following theorem will be proved that will be useful in further theorems.

Theorem 11. *Let P and Q be finite partitions of a quantum logic corresponding to a state s having Bayes’ Property. Then $$.*

*Proof.* Let *P = {a*_{1},..., a_{n}} and *Q = {b*_{1},..., b_{m}}. From Lemma 6 we can write

$$\begin{array}{l}{h}_{s}^{l}(P\vee Q)={\displaystyle \sum _{i=1}^{n}}{\displaystyle \sum _{j=1}^{m}}s({a}_{i}\wedge {b}_{j})(1-s({a}_{i}\wedge {b}_{j}))\\ ={\displaystyle \sum _{i=1}^{n}}{\displaystyle \sum _{j=1}^{m}}s({a}_{i}\wedge {b}_{j})-{\displaystyle \sum _{i=1}^{n}}{\displaystyle \sum _{j=1}^{m}}{(}^{s}\\ ={\displaystyle \sum _{j=1}^{m}s({b}_{j})-{\displaystyle \sum _{i=1}^{n}}{\displaystyle \sum _{j=1}^{m}}}{(}^{s}\end{array}$$*Hence the proof is complete.*

In the next theorem it is proved subadditivity of logical entropy of partitions on a quantum logic.

**Theorem 12.** *Let P and Q be finite partitions of a quantum logic corresponding to a state s having Bayes’Property. Then*

i)$$

ii)$$ max{*h*^{1}_{s}(P), h^{1}_{s}(Q)} ≤ *h*^{1}_{s}(P ∨ Q) ≤ h^{1}_{s}(P) + h^{1}_{s}(Q).

*Proof.* i) Let *P =* {a_{1},..., *a*_{n}} and *Q = {b*_{1},..., b_{m}}. For each *a*_{i} ε*P,* we can write

$$\begin{array}{l}{\displaystyle {\sum}_{j=1}^{m}}s({a}_{i}\wedge {b}_{j})(s{b}_{j})-s({a}_{i}\wedge {b}_{j}))\\ \le {\displaystyle {\sum}_{j=1}^{m}}s({a}_{i}\wedge {b}_{j})({\displaystyle {\sum}_{j=1}^{m}}s({b}_{j})-s({a}_{i}\wedge {b}_{j}))\\ =s({a}_{i})({\displaystyle {\sum}_{j=1}^{m}}s({b}_{j})-s({a}_{i}\wedge {b}_{j}))\\ =s({a}_{i})(1-{\displaystyle {\sum}_{j=1}^{m}}s({a}_{i}\wedge {b}_{j})=(s({a}_{i})(1-s({a}_{i})).\end{array}$$ii) From Theorem 11 and part i),

$$$$By Theorem 11,$$

Let s be a state. Two finite partitions *P* and *Q* of a quantum logic are called s-independent if *s(a ∧ b) = s(a)s(b)* for all *a ε P,* and *b ε Q.*

In the next theorem you observe that, for two *s*- independent finite partitions *P* and *Q* of a *QL, **h*^{1}_{s}(P ∨ Q).= *h*^{1}_{s}(P) + *h*^{1}_{s}(Q) necessarily. Also, in this case *h*^{1}_{s}(P|Q)= *h*^{1}_{s}(P) necessarily.

**Theorem 13**. *Let s be a state and let P and Q be s- independent finite partitions of a quantum logic. Then*

i)
$$$$;

ii)
$$$$

*Proof*. i) Since

$${h}_{s}^{l}(P)=1-{\displaystyle \sum _{i=1}^{n}}{(}^{s}\le 1-{\displaystyle \sum _{i=1}^{n}}{(}^{s}={h}_{s}^{l}(Q)$$ and *P, Q* are s-independent, we can write

$$\begin{array}{l}{h}_{s}^{l}(P\vee Q)=1-{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}{(}^{s}\\ 1-{\displaystyle {\sum}_{i=1}^{n}}{\displaystyle {\sum}_{j=1}^{m}}(s{(}^{{a}_{i}}{(}^{s}\\ 1-({\displaystyle {\sum}_{i=1}^{n}{(s({a}_{i}))}^{{}_{2}^{}}})({\displaystyle {\sum}_{j=1}^{m}{(s({b}_{j}))}^{{}_{2}^{}}})\\ 1-({\displaystyle {\sum}_{i=1}^{n}{(s({a}_{i}))}^{{}_{2}^{}}})({\displaystyle {\sum}_{j=1}^{m}{(s({b}_{j}))}^{{}_{2}^{}}})\\ -{\displaystyle {\sum}_{i=1}^{n}{(s({a}_{i}))}^{{}_{2}^{}}}+{\displaystyle {\sum}_{i=1}^{n}{(s({a}_{i}))}^{{}_{2}^{}}}\\ ={h}_{s}^{l}(P)+{\displaystyle {\sum}_{i=1}^{n}{(s({a}_{i}))}^{{}_{2}^{}}}(1-{\displaystyle {\sum}_{j=1}^{m}{(s({b}_{j}))}^{{}_{2}^{}}})\\ ={h}_{s}^{l}(P)+(1-{h}_{s}^{l}(P){h}_{s}^{l}(Q)\\ ={h}_{s}^{l}(P)+{h}_{s}^{l}(Q)-{h}_{s}^{l}(P){h}_{s}^{l}(Q)\end{array}$$ii) Follows from i) and Theorem 11. □

**Definition 14**. Let *P = {a*_{1},..., a_{n}} and *Q = {b*_{1},..., b_{m}} be two partitions of a quantum logic corresponding to a state s. We say *Q* is a s-rehnement of P, denoted by *P ≾*_{s} Q, if there exists a partition *I(1),..., I(n)* of the set *{1,..., m} *such that *a*_{i} = V_{jεI(i)}b_{j} for every *i = 1,..., n.*

Now the relation between the s-rehnement and the logical entropy of hnite partitions will be studied.

Theorem 15. *Let P = {a*_{1},..., a_{n}}, Q = {b_{1},..., b_{m}} and R = {c_{1},..., c_{r}} be partitions of a quantum logic corresponding to a state s. Then

i) P ≾_{s} Q implies that $$

ii) If P ≾_{s} Q and the quantum logic be distributive then $$

*Proof.* i)Since *P ≾*_{s} Q, there exists a partition *I*(1),..., *I*(n) of the set {1,...,*m*} such that *a*_{i} = V_{jεI(i)s}b_{j} for every *i = ..., n.* So from dehnition 2, *s(a*_{i}) = ∑_{jεI(i)}s(b_{j}), therefore $s({a}_{i})({\displaystyle {\sum}_{j=1}^{m}}s({b}_{j})-s({a}_{i}\wedge {b}_{j}))$ so

$${h}_{s}^{l}(P)=1-{\displaystyle \sum _{i=1}^{n}}{(}^{s}\le 1-{\displaystyle \sum _{i=1}^{n}}{(}^{s}={h}_{s}^{l}(Q)$$ii) P ≾_{s} Q implies that *P ∨ R ≾*_{s} Q ∨ R, because let *a*_{i} ∧ c be an arbitrary element of *P ∨ R,* then there exists a partition *I(1),..., I(n)* of the set {1,...,*m}* such that *a*_{i} = V_{jεI(i)}b_{j} for every *i* = 1,..., *n*. Therefore a_{i} ∧ c = (V_{jεI(i)}b_{j}) ∨ c = V_{jεI(i)}(b_{j} ∨ c), hence *P ∨ R ≾*_{s} Q ∨ R. Now by Theorems 11 and 15, it will be obtained $$

**Definition 16.** Let *P = {a*_{1},..., a_{n}} and *Q = {b*_{1},..., b_{m}} be two partitions of a quantum logic corresponding to a state s.$$ Q if for each *b*_{j} ε Q there exists *a*_{i} ε P with *s(a*_{i} ∧ b_{j}) = s(b_{j}). P ≗_{s} Q if *P ε*^{O}_{s} Q and *Q ε*^{O}_{s} P.

The next theorem shows that, the logical entropy and logical conditional entropy of hnite partitions of a quantum logic corresponding to a state s having Bayes’ Property, are invariant under the relation ≗_{s} .

**Theorem 17.** *Let P and Q be finite partitions of a quantum logic corresponding to a state s having Bayes’ Property. Then*

* i) P ε*^{O}_{s} Q if and only if h^{1}_{s}(P|Q) = 0;

*ii) if P ≗*_{s} Q then *h*^{1}_{s}(P) = *h*^{1}_{s}(Q);

*iii) if P ≗*_{s} Q then *h*^{1}_{s}(P|R) =*h*^{1}_{s}(Q|R);

* iv) if Q ≗*_{s} R then *h*^{1}_{s}(P|Q) = *h*^{1}_{s}(P|R).

*Proof.* i) Let *P = {a*_{1},..., a_{n}} and *Q = {b*_{1},..., b_{m}} and *P ε*^{O}_{s} Q, then for each *b*_{j} ε Q there exists *a*_{i0} ε P such that *s(b*_{j}) = s(a_{i0} Λ b_{j}). Since *s(b*_{j}) = ΣÌu *s(a*_{i} Λ b_{j}), we obtain *s(b*_{j}) = s(a_{i0} Λ b_{j}) and for each *i/ = j*_{0}, s(a_{i} ∧ bj) = 0 and so *h*^{1}_{s}(P|Q) = 0. Conversely, if *h*^{1}_{s}(P|Q) = 0 then for each *i, j, s(b*_{j}) = s(a_{i} ∧ b_{j}) or *s(a*_{i} ∧ b_{j}) = 0. For an arbitrary element *b*_{j} Q, since *0= s(b*_{j}) = ΣÌu *s(a; ∧ b*_{j}) we deduce that there exists an q, *1 ≤ i*_{1} ≤ n, such that *s(b*_{j}) = s(a_{i1} ∧ b_{j}). ii) Since *P ε*^{O}_{s} Q, by i) we have *h*^{1}_{s}(P|Q) = 0. So by theorem 11, *h*^{1}_{s}(P|Q) = h^{1}_{s} (P ∨ Q)- h^{1}_{s} (Q) = 0 and therefore * h*^{1}_{s} (P ∨ Q) = h^{1}_{s} (Q). Similarly if *Qε*^{O}_{s}P, then * h*^{1}_{s} (QIP) = * h*^{1}_{s} (P ∨ Q) - h’s(P) = 0 and so *h’s(P ∨ Q) = h’s(P).* Hence we imply that * h*^{1}_{s} (P) = h^{1}_{s} (Q).

iii) We first show that *Pε*^{O}_{s}Q implies that *P ∨ Rε*^{O}_{s}Q V R. Let *b*_{j}0 ∧ cko be an arbitrary element of *Q ∧ R.* Since *Pε*^{O}_{s}Q, there exists *aio e P* such that *s(a*_{io} ∧ b_{j0}) = s(b_{j0}). Now *s((aio ∧ ck0) ∧ (b*_{j0} ∧ c^)) = s(ai0 ∧ b_{j0} ∧ (ck0 ∧ ck0)) = s(aio ∧ b_{j0} ∧ ck0), it is sufficient to show that *s(aio ∧ b*_{j0} ∧ c_{k0}) = s(b_{j0} ∧c_{k0}). Since s has Bayes’ Property, *s(a*_{io} ∧b_{j0}) = s(b_{j}rj) = En=i *s(a*_{i} ∧ b_{j0}). So for each *i/ = i*_{o}, s(a_{i} ∧ b_{j0}) = 0, therefore for each *i/ = i*_{0}, s(a_{i} ∧ b_{j0} ∧ c_{k0}) = 0 and this implies that

$s({a}_{{i}_{o}}\vee {b}_{{j}_{o}}\vee {c}_{{k}_{o}})={\displaystyle \sum _{i=1}^{n}}s({a}_{{i}_{o}}\vee {b}_{{j}_{o}}\vee {c}_{{k}_{o}})=s({b}_{{j}_{o}}\vee {c}_{{k}_{o}}).$Thus *P ∨ Rε*^{O}_{s}Q ∧ R. By changing the role of *P* and *Q, P —s Q* implies that *P ∨ R ≗*_{s}Q ∧ R. Hence from ii), * h*^{1}_{s} (PR) = h^{1}_{s} (PVR)- h^{1}_{s} (R) = h^{1}_{s} (Q VR)- h^{1}_{s} (R) = h^{1}_{s} (QR). iv)Weneed toshowthat *Qε*^{O}_{s}R implies that *PVQε*^{O}_{s}PVR. Let *aio ∧cko *be an arbitrary element of *P ∨ R.* Since *Qε*^{O}_{s}R, there exists *b*_{j}o e Q such that *s(b*_{j}o ∧ ck0) = s(ck0). Now *s((ah ∧ b*_{j0}) ∧ (aio ∧ cko)) = s((aio ∧ aio) ∧ b_{j}o ∧ cko) = s(aio ∧ b_{j}o ∧ c_{ko}), it is sufficient to show that *s(aio ∧b*_{j0} ∧c_{ko}) = s(aio ∧c_{ko}). Since s has Bayes’ Property, *s(b*_{j}rj ∧ c_{ko}) = s(c_{ko}) = Eh s(b_{j} ∧ c_{ko}). So for each *j = j*_{0}, s(b_{j} ∧ c_{ko}) = 0, therefore for each *j = j*_{0}, s(a_{io} ∧ b_{j} ∧ c_{ko}) = 0 and this implies that

$$s({a}_{{i}_{o}}\wedge {b}_{{j}_{o}}\wedge {c}_{{k}_{o}})={\displaystyle \sum _{j=1}^{m}}s({a}_{{i}_{o}}\wedge {b}_{{j}_{o}}\wedge {c}_{{k}_{o}})=s({b}_{{j}_{o}}\wedge {c}_{{k}_{o}}).$$Thus *P ∨ Q c° P ∨ R.* By changing the role of *Q* and *R, Q* —s *R* implies that *P V Q —s P ∨ R.* Now from ii), * h*^{1}_{s} (P|Q) = * h*^{1}_{s} (P ∨ Q) - h^{1}_{s} (Q) = h^{1}_{s} (P ∨ R) - h^{1}_{s} (R) = h^{1}_{s} (PR). □

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