Please note in Fig. 1 that the 0, 1, 2, 3 rankings [please see J. C. Abbott, Sets, Lattices and Boolean Algebras, 1969, and D. E. Rutherford, Introduction to Lattice Theory, 1965] of L(30) -- for numbers with 0, 1, 2, 3 factors -- display a binomial-count pattern, fitting the 1, 3, 3, 1 row of Pascal's triangle (a general pattern abandoned by free lattices or arithmetics).
The Thesis will be validated by a combinatorial method perhaps unique to my work -- namely, "truth tables" or indicator tables for lattices or relational arithmetics. (They are, however, familiar in statement logic, set theory, and probability theory [Abbott,Rutherford].)
I claim this uniqueness because, conventionally in lattice theory [Abbott, Rutherford], the Sup is denoted by 1, the Inf as 0, confounding them with indicators wherein "1" denotes that a relation exists, "0" that a relation does not exist. I avoid confusion by denoting Supremum as "MAX", Infimum as "MIN".
Below, you see the indicator tables for factors of 30. Note that, if i(x) denotes tabular indicator of a relator, then i(LCM(a, b)) = MAX(i(a), i(b)) and i(GCD(a, b)) = MIN(i(a), i(b)).
1 2 3 5 6 10 15 30 0 0 0 0 0 0 0 0 0 0 0 1 0 1 1 1 0 0 1 0 1 0 1 1 0 0 1 1 1 1 1 1 0 1 0 0 1 1 0 1 0 1 0 1 1 1 1 1 0 1 1 0 1 1 1 1 1 1 1 1 1 1 1 1 (1) (4) (4) (4) (6) (6) (6) (7)I created a new measure: bank b counts the number of 1's in an indicator (column) -- in parentheses in the last row of Table.
Note that, theoretically for L(30), the bank could range 0-8, but banks 0, 2, 3, 5, 8 are missing from Table 1. Banks 1, 2, 3, 5 are recovered in F(30).
First, however, I briefly note a duality attribute of lattices. Lattice identities are invariant under the transformations of interchanging MAX with MIN, and JOIN with MEET [Abbott, Rutherford, and G. Birkhoff, Lattice Theory, 1967].
However, this is not a combinatorial duality. For example, ranks 2 and 3 have banks 6 and 7, a consecutive ordering, but ranks 2 and 1 (their dual transforms) have nonconsecutive banks 1 and 4. That is, LCM(2, 3), LCM(2, 5), LCM(3, 5), and LCM(2, 3, 5) are mutually nonequivalent; but GCD(2, 3) = GCD(2, 5) = GCD(3, 5) = GCD(2, 3, 5). This asymmetry is corrected in F(30).
However, there is no finite way of achieving a bank of 0 or a bank adding an indicator column of ALL 1's. These arise ideally (or axiomatically), as in set-theoretic topology. (Since these are the homologues of contradiction and truth, I believe this indicates the transfinitary nature of nonconstructive proofs.) No attempt will be made to recoup them in the free lattice. The finitary upper and lower bounds of the lattice will be denoted "MAX", "MIN".
Free structures are adorned with recondite elegance in the literature. Instead, I adapt a "basis" method (familiar in linear algebra) for constructing FREE lattices from lattices.
To the ATOMIC BASIS {2, 3, 5) for L(30), I adjoin MIN, for basis {1, 2, 3, 5}. I then
find that join, 
, applied to this extended basis yields L(30): {1, 2, 3, 5} = {1, 2, 3, 5, 6, 10, 15, 30}. Behold! L(30) is obtained merely from
JOIN! What of MEET?
When, guided by banks of the indicator tables, we apply meet, 
, we discover
F(30). (The term "free" means that we are free to apply any available operators or relators.)
Henceforth, I allow the abuse of language whereby, for "calculating", JOIN and MEET of elements
will, respectively, be denoted as MAX() and MIN() applied to their associated indicators; but,
for purposes of labeling FREE INDICATOR (TABLE 2), we shall use such forms as 2 3 or (2 3) (2 5),
etc.
From Table 1, find that MIN(i(2), i(3)) = 0, 0, 0, 0, 0, 0, 1, 1; that is, 2
3 has bank 2. Similarly, 2 5 has bank 2: MIN(i(2), i(5)) = 0, 0, 0, 0, 0, 1,
0, 1. Also 3 5: MIN(i(3), i(5)) = 0, 0, 0, 1, 0, 0, 0, 1.
What about JOINS of these newly obtained elements? We find that MAX(MIN(2, 3), MIN(2, 5)) =
i((2 3) (2 5)) = 0, 0, 0, 0, 0,
1, 1, 1, for b = 3; MAX(MIN(2, 3), MIN(3, 5)) = i((2 3)
(2 5)) = 0, 0, 0, 1, 0, 0, 1, 1; b = 3; MAX(MIN(2, 5), MIN(3, 5)) =
i((2 5) (3 5)) = 0, 0, 0, 1, 0, 1,
0, 1; b = 3.
We now have banks 1, 2, 3, 4, 6, 7. What of b = 5? We recoup this, and also a "new" element with b = 4.
Apply lattice duality to the results in the previous paragraph-but-one. Find that MIN(MAX(2, 3),
MAX(2, 5)) = i((2 3) (2 5)) =
i(6 10) = 0, 0, 0, 1, 1, 1, 1, 1, for b = 5; MIN(MAX(2, 3), MAX(3, 5)) =
i((2 3) (3 5)) = i(6 15) = 0, 0, 1, 1, 0, 1, 1, 1; b = 5; MIN(MAX(2, 5), MAX(3, 5)) = i((2 5) (3 5)) = i(10 15)
= 0, 1, 0, 1, 0, 1, 1, 1; b = 5.
We find one more result: MIN(MAX(2, 3), MAX(2, 5), MAX(3, 5)) = i((2 3)
(2 5) (3 5))
= i(6 10 15) = 0, 0, 0, 1, 0, 1, 1, 1, for b = 4. But
please note that this table is different from that for 2 and 3 and 5.
For compression, (2 3) (2 5) 
X; (2 3)
(3 5) Y;
(2 5) (3 5)
Z. We then have these results.
2Comparison of Table 2 with Table 1 shows that 9 of the 10 elements (10^15 is the exception) of Table 2 are factors or subdominants of 6, proving my Thesis. (The table for 6 is included in those tables.)
The Hasse diagram of F(30) appears in Fig. 2.
Note that F(30) possesses ranks 1, 2, 3, 4, 5, 6, 7, so that rank value of an element is its bank value. Note also that the distribution of elements over these rankings is 1, 3, 3, 4, 3, 3, 1, clearly not a binomial pattern. I know of no algorithm in the literature to determine the number of elements in F(n) from L(n).
Having discovered a way of generating free lattices, I was alerted for other instances of these structures. In a book on information retrieval (G. Salton, Automatic Information Organization and Retrieval, McGraw-Hill, 1968), I espied the Hasse diagram of L(30) as "description set" and its F(30) diagram as "request space". I believe that F(30) or, in general, F(n) should be used for both "spaces".
I said, above, that my Thesis invokes a famous debate between Niels Bohr and Albert Einstein. Einstein departed from relativistic thinking to argue that properties of elementary particles are attributive (inherent), a mode of thinking which may derive from ancient animistic religious beliefs. Bohr argued that these (relativistically) depend upon the given context.
I've found a contextual aspect for F(30). The indicator tables for L(6) (whose Hasse diagram can be deduced from the corresponding sublattice in L(30) are
1 2 3 6 0 0 0 0 0 0 1 1 0 1 0 1 1 1 1 1 (1) (2) (2) (3)Note that, in contrast to the tables for L(30) (Table 1), the banks are in consecutive order. You will also find that you cannot derive a properly free element. Even if you form the extended basis {1, 2, 3} and derive L(6) as V{1, 2, 3}, applying meet to the above tables yields nothing new. Just as noncommutativity of rotation arises only in passing from 2-D to 3-D, so properly free elements arise only in passing from two to three or more atomic elements. Hence L(6) has only two proper factors or dominants. But you readily see, from Fig. 2, that embedding L(6) in L(30) and "freeing" the latter results in 6 having 10 proper factors or dominants -- a contextual result.
This suggests a further analogy. A free quanton behaves like a wave; a bound quanton behaves likes a particle. Nonembedded 6 behaves according to the fundamental theorem of natural number arithmetic; embedded, 6 does not.
When L(6) is embedded in L(210), to obtain F(210), 6 has 24 factors or subdominants. As the number of atoms (or primes) increases, the number of factors or subdominants of a given element increases monotonically. In terms of (nonfree) relational databases, this indicates millions of unasked questions! Or probabilities of events!
(RELATIONAL ARITHMETIC can be formulated in terms of "Pythagorean dots", making it INDEPENDENT OF OPERATIONAL ARITHMETIC.)