There are two definitions of a partially ordered set (abbreviated to 'poset'), each having its own advantages and disadvantages.

DEFINITION I. Let R be a binary relation on the set A such that for any
elements a, b, c of A,

1. aRa 2. aRb & bRa ==> a=b 3. aRb & bRc ==> aRc

DEFINITION II. Let < be a relation on the set A such that for any
elements a, b, c of A,

i. not a<a ii. a<b & b<c ==> a<c

LEMMA 1. Definitions I, II agree if R is '< or ='.

*Proof*. We first show that Defn. II implies Defn. I if R is
interpreted as '< or ='. Clearly 1 holds because a=a. For 3, if aRb
& bRc, then either a=b=c (so a=c), or a=b<c, or a<b=c (both giving a<c),
or a<b<c (so a<c by ii). Finally, 2 holds because its left hand side
tells us that either a<b & b<a (giving a<a by ii; which
contradicts i), or a=b & b<a, or a<b & b=a (both giving a<a
again), or a=b=a as desired.

Now we show that Defn. I implies Defn. II. Given the relation R, we define
the relation < on A by

(1) a<b iff aRb & a=/=b. Hence aRb iff a<b or a=b.

To prove i: the negative of (1) gives

(not aRb or a=b) ==> not a<b.

Putting b=a enables us to argue:

a=a ==> (not aRa or a=a) ==> not a<a.

For ii:

a<b & b<c ==> a<b & b<c & (a=c or a=/=c) ==> (aRb & bRc & b=/=a,c & a=c) or (aRb & bRc & b=/=a,c & a=/=c) ==> (aRb & bRa & b=/=c & a=c) or (aRc & a=/=c) ==> (b=a & b=/=c & a=c) or a<c ==> a<c.

Last updated: 04 January 2005