Formulated 2-Traces

The formulation f(T) of a trace T=t₁,...t_n over $\Sigma$ is the Boolean function $\lor(\land T (t_i \in T))$ which forms the conjunction of elements of a 2-trace and then disjoins the results. For example, if $T^2_3 (\alpha, \beta, \gamma) = \{\{\alpha,\beta\},\{\alpha,\gamma\},\{\beta,\gamma\}\}$ then  

$$f(T^2_3) = (\alpha \land \beta) \lor (\alpha \land \gamma) \lor (\beta \land \gamma) = 2/3(\alpha, \beta, \gamma).$$ (1)

The significance of formulated traces is that, if a sentence $\alpha$follows classically from every element of a trace T, then $f(T)\rightarrow\alpha$ is a theorem of classical logic.

To illustrate the dependence of least fundamental trace members on n, notice that the least fundamental traces T²_n for n = 3, 4 and 5 and inputs $\Sigma = \{a,b,c,d,e\}$ are:

$$T^2_3(a,b,c) = \{\{ab\},\{ac\},\{bc\}\}$$

$$T^2_4(a,b,c,d) = \{\{ab\},\{ac\},\{ad\}, \{bcd\}\}$$

$$T^2_5(a,b,c,d,e) = \{\{ab\},\{ac\},\{bcd\}, \{bce\}, \{ade\}\}$$

T²_n for any arbitrary n atoms, $\{a,b,c,\cdots,\alpha_n\}$can be expressed as a recursive function  

$$T^2_n = f(T^2_3)(T^2_{n-1},T^2_{n-2},\alpha_n).$$ (2)

Apostoli and Brown prove in [1] that a set of sentences closed under 2/(n + 1) including single elements, contains all formulated n-traces. The question is whether there is an alternative proof of this result that relies upon visualization techniques. A visual approach would help in highlighting features in traces that give rise to large numbers of permutations. An algebraic description of the set of n-traces and a graphic representation would give a precise account of those permutations.