Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am reading Part II of Chin and Tarski's "Distributive and Modular Laws in the Arithmetic of Relation Algebras". In the beginning of section 4, the authors say "In general, if $\odot$ is a binary operation which is commutative, associative, and tautological, then both the left and right distributive laws for $\odot$ under $\odot$ are identically satisfied". My question is: what does "tautological" mean in this context?

We are working within a fixed Boolean algebra $B$. $\odot$ is some binary operation on the universe of $B$. He is saying that if $\odot$ is commutative, associative, and "tautological", then $\odot$ is left and right distributive over $\odot$, i.e., for all $x$, $y$, and $z$ in the universe of $B$, $x\odot (y\odot z)=(x\odot y)\odot (x\odot z)$ and $(y\odot z)\odot x=(y\odot x)\odot (z\odot x)$. I can't find the word "tautological" (in reference to a binary relation) used anywhere else.

share|improve this question
    
Do they use the term anywhere else? Can you give us any more of the text, or context here? –  Doug Spoonwood Mar 27 '12 at 1:31
    
Could it possibly mean idempotent? –  Miha Habič Mar 27 '12 at 5:44

1 Answer 1

up vote 3 down vote accepted

Yes, it means idempotence (for binary operations). A binary operation $\odot$ is tautological iff $\forall x\ [x \odot x = x]$. Perhaps the name comes from the fact that conjunction and disjunction satisfy this tautologically. I think this is old terminology, but I personally find it somewhat appealing to reserve idempotence for unary operations since the idea is somewhat different. Anyway, here's an example, where it's called the law of tautology (pg. 211):

Frink, Orrin. New Algebras of Logic. The American Mathematical Monthly , Vol. 45, No. 4 (Apr., 1938), pp. 210-219 http://www.jstor.org/stable/2302605

share|improve this answer
    
Thank you, Rachel :). –  MathMastersStudent Mar 28 '12 at 2:11

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.