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+2 votes
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asked in Set Theory & Algebra by Active (1.9k points)  
recategorized by | 161 views

4 Answers

+3 votes

Option a

As complement

m∨n=1 and m∧n=0

m∨p=1 and m∧p=0

So both n and p are complements of m

For distirbutive lattice

N∧(m∨p)= (N∧m)∨(N∧p)

From lhs.    n∧(m∨p)=n∧1=n

From rhs. n∧m=0     n∧p=p

And     0∨p= p

So lhs≠ rhs that's why given lattice is non-distributive

answered by Boss (5.7k points)  
edited by
+1 vote
Answer will be (A)

S1: is true because complement of a lattice and lattice has a lub and a glb

S2: Distributive lattice each element has atmost one complement. but here m has 2 complement
answered by Veteran (52.4k points)  
0 votes

S1: for lattice to be complemented there should exist LUB and GLB for pair of elements 
eg :(m,n) its LUB is 1 and its GLB is 0
 S2: for a lattice to be distributive complement should be unique in above diagram we have two complements of m that is n,p
so it is not distributive lattice 
so  (A) option 

answered by Veteran (19.9k points)  
0 votes
By above diagram you can see that complement of m is n and p.

Since there exist an vertex m such that it have more than on complement so it voilet the deffinition of distributive lattice.
answered by Loyal (4.5k points)  


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