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## Here's the Solution to this Question

Since, given statement is unclear, we assume we need to show that:

A lattice L is distributive if and only if

$(a \vee b) \wedge(b \vee c) \wedge(c \vee a)=(a \wedge b) \vee(b \wedge c) \vee(c \wedge a) \forall a, b, c \in L$ .

### Proof: We suppose that L is distributive lattice.

\begin{aligned}(a v b) \wedge(b \vee c) \wedge(c \vee a) &=a \wedge[(b \vee c) \wedge(c \wedge a)]\} \vee\{b \wedge[(b \vee c) \wedge(c \vee a)]\} \\ &=[(a \wedge(c \wedge a) \wedge(b \vee c)] \vee[(b \wedge(b \vee c)) \wedge(c \vee a)]\\ &=[(a \wedge(b \vee c)] \vee[b \wedge(c \vee a)]\\ &=[(a \wedge b) \vee(a \wedge c)] \vee[(b \wedge c) \vee(b \wedge a)] \\ &=(a \wedge b) \vee(b \wedge c) \vee(c \wedge a) \end{aligned}

Conversely, we first show that L is modular.

Let x, y, z be any three elements of L with $x \leqslant z$

\begin{aligned} x \vee(y \wedge z) &=[x \vee(x \wedge y)] \vee(y \wedge z) \text { [by absorption] } \\ &=(x \wedge z) \vee(x \wedge y) \vee(y \wedge z) \\ &=(x \vee y) \wedge(y \vee z) \wedge(z \vee x) \\ &=(x \vee y) \wedge[(y \vee z) \wedge z] \quad[\because x \wedge z=x] \\ &=(x \vee y) \wedge z \quad[\text { by absorption property }] \end{aligned} \\ =(x \vee y) \wedge[(y \vee z) \wedge z] \quad[\because x \leqslant z]

$=(x \vee y) \wedge z \quad$ [by absorption property]

Thus, L is modular.

\begin{aligned} &\text { Now for any } a, b, c \in L \\ &\qquad \begin{aligned} a \wedge(b \vee c) &=[a \wedge(a \vee c)] \wedge(b \vee c) \\ &[b y \text { absorption property }] \\ &=[(a \vee b) \wedge(b \vee c) \wedge(c \vee a)] \wedge a \\ &=[(a \wedge b) \vee(b \wedge c) \vee(c \wedge a)] \wedge a \end{aligned} \\ &=[(a \wedge b) \vee(c \wedge a)) \vee(b \wedge c)] \wedge a \\ &\text { Since } \\ &\begin{aligned} a \wedge b \leqslant a, a \wedge c \leqslant a & \Rightarrow(a \wedge b) \vee(a \wedge c) \leqslant a, \\ \text { we can apply modular identity on R.H.S of }(1) \text { to get } \\ a \wedge(b \vee c) &=[(a \wedge b) \vee(c \wedge a)] \vee[(b \wedge c) \wedge a] \\ &=(a \wedge b) \vee[(c \wedge a) \vee(b \wedge(c \wedge a))] \\ &=(a \wedge b) \vee(c \wedge a) \end{aligned} \\ &\qquad \begin{aligned} L \text { is distributive. } &[\text { by absorption property }] \end{aligned} \end{aligned} Hence, proved.