Show that a chain is a distributive lattice I made this proof, but I don't know if it's correct. Someone correct me if Im wrong.
Showing a chain $(P,\leqslant)$ is a distributive lattice:
Assuming $x \leqslant y \leqslant z$ , for $x,y,z \in P$ 
$(L1)$
We have $x \wedge y =x=y \wedge x.$ Also , $x \vee y =y=y \vee x.$ Easily we see that the comutative property holds for all possible relations between $x,y,c$.
$(L2)$
We have $x \wedge (y \wedge z)=x \wedge y =x = x \wedge z= (x \wedge y)  \wedge z$.
Also, $x \vee (y \vee z)=x \vee z=z=y \vee z=(x \vee y ) \vee z$. So associative property also holds.
$(L3)$ We have $x \wedge ( x \vee y)= x \wedge y=x$, and , $x \vee ( x \wedge y)=x \vee x=x.$
$(L4)$ (idempotent) Also verifies, because $L1$,$L2$,$L3$ verify
Therefore P is a lattice.
To proove its a distributive lattice, I just need to verify that in every possible relation between $x,y,z \in P$ the distributive property holds.
Like $x \wedge(y \vee z)=x \wedge z=x= x \vee x = (x \wedge y) \vee (x \wedge z)$
 A: To construct a lattice out of a chain (K,<=), first define
. . . x min y = x if x <= y, = y if y <= x,
. . . x max y = y if x <= y, = x if y <= x.  
Since K is a chain, these definitions are well defined for all x,y.  
To show x min y is the greatest lower bound of A = { x,y },
assume b is a lower bound of A.  Thus b <= x,y and either
x <= x min y or y <= x min y.  Either way b <= x min y.
Thus x min y is the glb of A.  
In a similar manner, x max y is the least upper bound of A.
Consequently (K,min,max) is a lattice.  
Since a = a min b iff a <= b iff b = a max b
one can, if desired, prove all the equations
for an algebraic description of a lattice.  
To show (K,min,max) is a distributive lattice both
. . a min (x max y) = (a min x) max (a min y)
and
. . a max (x min y) = (a max x) min (a max y)
are to be proven.  
Before proceeding with the proof, show that if x <= y,
then a min x <= a min y and a max x <= a max y.  
For the first equation, there are two cases:  x <= y and y <= x
which by symmetry, only one needs to be considered, that is wlog.  
So assume x <= y.  Whence
(a min x) max (a min y) = a min y = a min (x max y).  
The second equation is handled in a similar manner.  
