Leave $ s in Sigma ^ * $. Leave elements of the finite set $ A $ be collections of subintervals $ (i, j) $ from $ (0, | s |) $. Such that yes $ x in A $ Y $ y, z in x $ then also $ y subset z, z subset and $ or $ y cap z = {} $. Clearly, this captures the idea of a chain grouping $ s $:

$$

((abc) d) f (a (bc) d) e

$$

through parentheses. It is also obvious that you can convert the previous grouping into a CFG generator $ s $:

$$

S to AfBe \

A a Cd \

B a aDd \

C a abc

$$

and vice versa. Therefore, the generation of CFG $ s $ correspond one by one with valid parenthesis groupings of $ s $. Grouping notation is easier to work than grammar notation.

Now define $ (i, j) cap (a, b) = ( max (i, a), min (j, b)) $ where if the result is $ (l, k) $ Y $ k leq l $ we say that they don't cross or that their intersection is $ 0 $.

Now for any $ x, y in A $ define $ x cdot y = {a cap b: a in x, b in y } $. This operation forms a monoid through the associativity of $ cap $ giving associativity of the intersection of elements, and identity is the element $ {(0, | s |) } in A $. We will refer to that element as $ 1 $. There is a zero element, namely $ {} $ which is multiplied by anything to match $ {} $.

Henceforth, $ (i, j) equiv {(i, j) } in A $ with notation

If we define the union of two singleton elements $ (i, j), (a, b) in A $ be $ {(i, j), (a, b) } $ Yes $ i neq b $ Y $ j neq to $ Y $ (i, j) cap (a, b) = 0 $, and if they touch or overlap $ (i, j) cap (a, b): = ( min (i, a), max (j, b)) $ (its union in the rope $ s $); so **Can we have a ring using some kind of symmetric difference or element by element?**

Note that the power set of a set forms a ring below $ + = $ symmetric difference

Also, keep in mind that I have posted about this ring before, but I ran into problems that don't have singleton junctions two joints if they touch or overlap.

If you need to define the plugin first, say if you were trying $ x + y = x overline {y} cup overline {x} y $ trick, then obviously yes $ (i, j) $ it's a singleton in $ A $, its complement would be:

$$

{(0, i), (j, | s |) } in A

$$

It is very confusing for me how to extend the complement to other elements, etc. Please try it.