Since, given statement is unclear, we assume we need to show that:
A lattice L is distributive if and only if
(a∨b)∧(b∨c)∧(c∨a)=(a∧b)∨(b∧c)∨(c∧a)∀a,b,c∈L .
(avb)∧(b∨c)∧(c∨a)=a∧[(b∨c)∧(c∧a)]}∨{b∧[(b∨c)∧(c∨a)]}=[(a∧(c∧a)∧(b∨c)]∨[(b∧(b∨c))∧(c∨a)]=[(a∧(b∨c)]∨[b∧(c∨a)]=[(a∧b)∨(a∧c)]∨[(b∧c)∨(b∧a)]=(a∧b)∨(b∧c)∨(c∧a)
Conversely, we first show that L is modular.
Let x, y, z be any three elements of L with x⩽z
x∨(y∧z)=[x∨(x∧y)]∨(y∧z) [by absorption] =(x∧z)∨(x∧y)∨(y∧z)=(x∨y)∧(y∨z)∧(z∨x)=(x∨y)∧[(y∨z)∧z][∵x∧z=x]=(x∨y)∧z[ by absorption property ]=(x∨y)∧[(y∨z)∧z][∵x⩽z]
=(x∨y)∧z [by absorption property]
Thus, L is modular.
Now for any a,b,c∈La∧(b∨c)=[a∧(a∨c)]∧(b∨c)[by absorption property ]=[(a∨b)∧(b∨c)∧(c∨a)]∧a=[(a∧b)∨(b∧c)∨(c∧a)]∧a=[(a∧b)∨(c∧a))∨(b∧c)]∧a Since a∧b⩽a,a∧c⩽a we can apply modular identity on R.H.S of (1) to get a∧(b∨c)⇒(a∧b)∨(a∧c)⩽a,=[(a∧b)∨(c∧a)]∨[(b∧c)∧a]=(a∧b)∨[(c∧a)∨(b∧(c∧a))]=(a∧b)∨(c∧a)L is distributive. [ by absorption property ] Hence, proved.