algebraic definition of a lattice
The parent entry (http://planetmath.org/Lattice^{}) defines a lattice as a relational structure (a poset) satisfying the condition that every pair of elements has a supremum^{} and an infimum^{}. Alternatively and equivalently, a lattice $L$ can be a defined directly as an algebraic structure^{} with two binary operations^{} called meet $\wedge $ and join $\vee $ satisfying the following conditions:

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(idempotency of $\vee $ and $\wedge $): for each $a\in L$, $a\vee a=a\wedge a=a$;

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(commutativity of $\vee $ and $\wedge $): for every $a,b\in L$, $a\vee b=b\vee a$ and $a\wedge b=b\wedge a$;

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(associativity of $\vee $ and $\wedge $): for every $a,b,c\in L$, $a\vee (b\vee c)=(a\vee b)\vee c$ and $a\wedge (b\wedge c)=(a\wedge b)\wedge c$; and

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(absorption): for every $a,b\in L$, $a\wedge (a\vee b)=a$ and $a\vee (a\wedge b)=a$.
It is easy to see that this definition is equivalent^{} to the one given in the parent, as follows: define a binary relation^{} $\le $ on $L$ such that
$$a\le b\mathit{\hspace{1em}}\text{iff}\mathit{\hspace{1em}}a\vee b=b.$$ 
Then $\le $ is reflexive^{} by the idempotency of $\vee $. Next, if $a\le b$ and $b\le a$, then $a=a\vee b=b$, so $\le $ is antisymmetric. Finally, if $a\le b$ and $b\le c$, then $a\vee c=a\vee (b\vee c)=(a\vee b)\vee c=b\vee c=c$, and therefore $a\le c$. So $\le $ is transitive^{}. This shows that $\le $ is a partial order^{} on $L$. For any $a,b\in L$, $a\vee (a\vee b)=(a\vee a)\vee b=a\vee b$ so that $a\le a\vee b$. Similarly, $b\le a\vee b$. If $a\le c$ and $b\le c$, then $(a\vee b)\vee c=a\vee (b\vee c)=a\vee c=c$. This shows that $a\vee b$ is the supremum of $a$ and $b$. Similarly, $a\wedge b$ is the infimum of $a$ and $b$.
Conversely, if $(L,\le )$ is defined as in the parent entry, then by defining
$$a\vee b=sup\{a,b\}\mathit{\hspace{1em}}\text{and}\mathit{\hspace{1em}}a\wedge b=inf\{a,b\},$$ 
the four conditions above are satisfied. For example, let us show one of the absorption laws: $a\vee (a\wedge b)=a$. Let $c=inf\{a,b\}\le a=a\wedge b$. Then $c\le a$ so that $sup\{a,c\}=a$, which precisely translates to $a=a\vee c=a\vee (a\wedge b)$. The remainder of the proof is left for the reader to try.
Title  algebraic definition of a lattice 

Canonical name  AlgebraicDefinitionOfALattice 
Date of creation  20130322 17:39:29 
Last modified on  20130322 17:39:29 
Owner  CWoo (3771) 
Last modified by  CWoo (3771) 
Numerical id  12 
Author  CWoo (3771) 
Entry type  Definition 
Classification  msc 03G10 
Classification  msc 06B99 