JAIST seminar talk

<img src=../../PCP/Robert_Dilworth1.jpg>
 

</center>

Residuated lattice-ordered monoids (residuated lattices) are algebras of the form
$\m L=\<L,\vee,\wedge,\cdot, e,\ld,\rd\>$ such that
<blockquote>
(L)   $\<L,\vee,\wedge\>$ is a lattice

(M)   $\<L,\cdot,e\>$ is a monoid, i.e., $\cdot$ is
associative and $x\cdot e=x=e\cdot x$

(R)   $xy\le z$ iff $x\le z\rd y$ iff $y\le x\ld z$ (for all $x,y,z\in L$)

These equivalences are equivalent to identities. E.g. we can take
 
$x(x\ld z\wedge y)\le z$ and $y\le x\ld (xy\vee z)$ $(*)$

and the mirror images of these identities. 
</blockquote>

Note that the operation $\cdot$ has priority over $\ld,\rd$, and they
are performed before $\vee$, $\wedge$.

The two inequalities $(*)$ follow directly from the equivalence
$xy\le z\ \text{ iff } y\le x\ld z.$

Conversely, if the inequalities $(*)$ hold, then 

$xy\le z$ implies $y\le x\ld (xy\vee z)=x\ld z$, and 

$y\le x\ld z$ implies $xy=x(x\ld z\wedge y)\le z$. 

Hence the identities indeed capture the property of being residuated.

Conclusion: $\m{RL}=\{$all residuated lattices$\}$ is a variety.

The argument just given is a special case of the following simple but useful lemma.

Lemma: Let $r,s,t$ be terms in a (semi)lattice-ordered algebra, and
denote the sequence of variables $y_1,\ldots,y_n$ by $\m y$. Then the quasi-identity
<center>
$x\le r(\m y)\Rightarrow s(x,\m y)\le t(x,\m y)$
</center>
is equivalent to the identity
<center>
$s(x\wedge r(\m y),\m y)\le t(x\wedge r(\m y),\m y)$.
</center>
<a href=qe2eq.html>Interactive Proof Puzzle (requires Internet Explorer)</a>

In general, quasi-identities are not preserved by homomorphic images. E.g. any free monoid
satisfies $xz=yz\Rightarrow x=y$ but has the non-cancellative monoid $\{e,0\}$ as
homomorphic image.

The lemma above shows that certain quasi-identities are preserved by homomorphic
images. Combined with the property of residuation, it allows such quasi-identites
to be translated to identities. 

E.g. $xz\le yz\Rightarrow x\le y$ is equivalent to $x\le yz\rd z\Rightarrow x\le y$,
which translates to $x\wedge yz\rd z\le y$ (and simplifies to $yz\rd z=y$).

Together with its mirror image, we have proved that in residuated lattices cancellativity
can be expressed by identities.

On Wednesday I will present a more sophisticated result that shows how any
positive universal sentence $\sigma$ can be translated to set of identities $E(\sigma)$ 
that is equivalent to $\sigma$ in all subdirectly irreducible residuated lattices.

Note that we do not assume that $\m L$ is bounded or commutative or that $e$ is
the top element (otherwise we could not have the following result).

Theorem $\m{LG}$ is term equivalent to a subvariety of $\m{RL}$.

Defining $\ld,\rd$ is an $l$-group is easy: $x\ld y=x^{-1}y$ and $x\rd y=xy^{-1}$.

Exercise 7: prove that if $\m G$ is an $l$-group, then $\<G,\vee,\wedge,\cdot,e,\ld,\rd\>$ is a
residuated lattice that satisfies $x(x\ld e)=e$.

Defining $^{-1}$ in a residuated lattice is equally simple: $x^{-1}=x\ld e$.

Exercise 8: prove that if $\m L$ is a residuated lattice that satisfies $x(x\ld e)=e$, 
then $\<L,\vee,\wedge,\cdot,e,^{-1}\>$ is an $l$-group, and $\m L$ satsifies
$(x\ld e) y=x\ld y$, $x(y\ld e)=x\rd y$.

Corollary The lattice of $l$-group subvarieties is (isomorphic to) an ideal in 
the lattice of residuated lattice subvarieties.

On Wednesday we will see how this result can even illuminate another part of the lattice
of subvarieties of $\m{RL}$.