We have proved that irreducible 2nd dimensional submodules/quotient modules from the module of reducible Appell F1 functions comes from 2F1.

Now turn to another 3rd order A-hypergeometric function G3(a,b,x,y) and try to get similar results.

First, we look up in the literature for reducible condition of G3(a,b,x,y): reducible iff 2*a+b or a+2*b is an integer.

STEP 1. Pick three linearly-independent elements in the system as a   
        basis for C(x,y)[Dx,Dy]-module of G3 and find the expression of 
        this module by solving equations. Basis: G3, Dx(G3), Dy(G3) 
        (module is much bigger expression than module of F1).

STEP 2. Apply "hom" on reducible modules of G3, we found there are only 
        4 cases left: a+2*b=0; a+2*b=1; 2*a+b=0; 2*a+b=1.

STEP 3. Choose one case: a+2*b=1, i.e., look into the module of G3(-2*b+
        1,b,x,y). Now we want to get a formula of G3(-2*b+1,b,x,y)
         =...2f1(A,B,C,F) where parameters A,B,C only depends on b and F  
         is the change of variable function in C(x,y). 

First, we uncover F. Fix b with some value. Then substitute y with some function in x (use "changeVmoduleF1"), now we have the module of G3(-2*b+1,b,x,f(x)) with b being some value. Then find the 3rd order differential operator of this module (use "CycVec"), and then factor it and get the 2nd order differential operator (use "DFactor" to get [order 1, order 2] factorization), then use Erdal's program "find_2f1" to find its 2f1 solution. Collect a bunch of such y and change of variable function in the 2f1 solution, then combine them to obtain the formula for F.

STEP 4. Check if F is correct and also find the formula for A,B,C in 
        2f1. (checkPullback.txt)

STEP 5. Now we have formula for parameters in 2f1: A,B,C and change of 
        variable function F. We can check if the module of G3(-2*b+
        1,b,x,y) comes from 2f1(A,B,C,F) (use "homSurj" on them). 

STEP 6. There exist projective homomorphisms between G3 and 2f1, so 2nd 
         order submodule or quotient module of G3(-2*b+1,b,x,y) comes 
         from 2f1.