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[parent] distributive inequalities (Derivation)

Let $L$ be a lattice. Then for $a,b,c\in L$ , we have the following inequalities:

  1. $a\vee (b\wedge c)\le (a\vee b)\wedge (a\vee c)$ ,
  2. $(a\wedge b)\vee (a\wedge c)\le a\wedge (b\vee c)$ .
Proof. Since $a\le a\vee b$ and $a\le a\vee c$ , $a\le (a\vee b)\wedge (a\vee c)$ . Similarly, $b\wedge c \le b\le a\vee b$ and $b\wedge c\le c\le a\vee c$ imply $b\wedge c\le (a\vee b)\wedge (a\vee c)$ . Together, we have $a\vee (b\wedge c)\le (a\vee b)\wedge (a\vee c)$ .

The second inequality is the dual of the first one. $ \qedsymbol$

The two inequalities above are called the distributive inequalities.

Proposition A lattice $L$ is a distributive lattice if one of the following inequalities holds:

  1. $(a\vee b)\wedge (a\vee c)\le a\vee (b\wedge c)$ ,
  2. $a\wedge (b\vee c)\le (a\wedge b)\vee (a\wedge c)$ .
Proof. By the distributive inequalities, all we need to show is that 1. implies 2. (that 2. implies 1. is just the dual statement). So suppose 1. holds. Then
$\displaystyle (a\wedge b)\vee (a\wedge c)$ $\displaystyle \ge ((a\wedge b)\vee a)\wedge ((a\wedge b)\vee c)$    by assumption    
  $\displaystyle = a \wedge ((a\wedge b)\vee c)$    by absorption    
  $\displaystyle \ge a\wedge ((c\vee a)\wedge (c\vee b))$    by assumption    
  $\displaystyle = (a\wedge (c\vee a))\wedge (c\vee b)$    meet associativity    
  $\displaystyle = a \wedge (c\vee b).$    by absorption    

$ \qedsymbol$



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See Also: modular inequality


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Cross-references: distributive lattice, proposition, imply, inequalities, lattice
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This is version 2 of distributive inequalities, born on 2007-01-27, modified 2007-05-04.
Object id is 8830, canonical name is DistributiveInequalities.
Accessed 1077 times total.

Classification:
AMS MSC: 06D99 (Order, lattices, ordered algebraic structures :: Distributive lattices :: Miscellaneous)
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