algorithm for generating ℱ[S]

Needs["DiscreteMath`Combinatorica`"]
P1[S_] := 
      Flatten[Join[Table[KSubsets[S, i], {i, 1, Length[S]}]], 1]; 
P2[S_] := 
      Flatten[Permutations /@ 
          Select[KSubsets[P1[S], 2], 
            Length[(#1[[1]]) ⋂ (#1[[2]])] == 
                0 &], 1]; 
𝔣[{s1_, s2_}] := F[And @@ s1, And @@ s2]; 
ℱ[S_] := 𝔣 /@ P2[S];

examples

ℱ[{p, q}] // TraditionalForm

ℱ[{p, q, r}] // TraditionalForm

ℱ[{p, q, r, s}] // TraditionalForm

Table[Length[ℱ[Range[n]]], {n, 2, 10}]

Adding T to a (contingent) coherent set of size 2 can never yield an incoherent set

Here's  ℱ[{p,q,T}]:

ℱ[{p, q, T}] // TraditionalForm

Any term of the form F[_, T] will be zero, by the definition of F.  So,

% //. F[x_, T] -> 0 // TraditionalForm

Moreover, any conjunction of the form X & T can be rewritten as X, since it is logically equivalent to X:

% //. x_ && T -> x // TraditionalForm

Finally, any term of the form F[T,_] will be 1, by the definition of F:

% //. F[T, x_] -> 1 // TraditionalForm

Now, we can take the Mean, to yield 𝒞({p,q,T}):

Mean[%] // TraditionalForm

Now, let's compare this with 𝒞({p,q}):

Mean[ℱ[{p, q}]] // TraditionalForm

The following proves that If 𝒞({p,q}) ≥ 0, then 𝒞({p,q,T}) ≥ 0.

Needs["Algebra`InequalitySolve`"]

InequalitySolve[
      (x/2) ≥ 0 && 
        (1/12) (3 x + 3) < 0 && -2 ≤ x ≤ 
          2, {x}]

Adding T to a  (contingent) coherent set of size 3 can never yield an incoherent set

ℱ[{p, q, r, T}] // TraditionalForm

% //. F[x_, T] -> 0 // TraditionalForm

% //. x_ && T -> x // TraditionalForm

% //. F[T, x_] -> 1 // TraditionalForm

Mean[%] // TraditionalForm

Mean[ℱ[{p, q, r}]] // TraditionalForm

The following proves that if 𝒞({p,q,r}) ≥ 0, then 𝒞({p,q,r,T}) ≥ 0.

Needs["Algebra`InequalitySolve`"]

InequalitySolve[
      (x/12) ≥ 0 && 
        (1/50) (3 x + 7) < 0 && -12 ≤ x ≤ 
          12, {x}]

This can easily be generalized to an inductive proof, for all n.

But, we can still have the coherence being DECREASED when we add a tautology.  This is seen in the following result (n = 2 case):

InequalitySolve[
      (x/2) > (1/12) (3 x + 3) && -2 ≤ x ≤ 
          2, {x}]

This is an artifact of the averaging in the definition.  If we just take the sum of the F-values, then this cannot happen.  In the n = 2 case:

𝒞({p, q})  =  F(p, q) + F(q, p)

𝒞({p,q,T}) = 3  F (p, q) + 3 F(q, p)  +  3  =  
          3 [𝒞({p, q})  +  1]