YES

Problem:
 F(G(x,A(),B())) -> x
 G(F(H(C(),D())),x,y) -> H(K1(x),K2(y))
 K1(A()) -> C()
 K2(B()) -> D()

Proof:
 Matrix Interpretation Processor:
  dimension: 3
  interpretation:
              [1 0 0]     [0]
   [K2](x0) = [0 1 1]x0 + [0]
              [0 0 0]     [1],
   
              [1 1 0]  
   [K1](x0) = [0 0 1]x0
              [0 0 0]  ,
   
                 [1 1 0]     [1 1 0]  
   [H](x0, x1) = [0 0 0]x0 + [1 0 0]x1
                 [0 1 0]     [0 0 1]  ,
   
         [0]
   [D] = [0]
         [0],
   
         [0]
   [C] = [0]
         [0],
   
             [1 0 0]     [0]
   [F](x0) = [0 0 1]x0 + [1]
             [0 1 1]     [0],
   
                     [1 1 1]     [1 1 1]     [1 1 1]     [0]
   [G](x0, x1, x2) = [0 0 0]x0 + [0 0 0]x1 + [1 0 0]x2 + [1]
                     [1 1 1]     [0 1 1]     [0 0 0]     [0],
   
         [0]
   [B] = [0]
         [0],
   
         [1]
   [A] = [0]
         [0]
  orientation:
                     [1 1 1]    [1]         
   F(G(x,A(),B())) = [1 1 1]x + [1] >= x = x
                     [1 1 1]    [1]         
   
                          [1 1 1]    [1 1 1]    [1]    [1 1 1]    [1 1 1]    [0]                 
   G(F(H(C(),D())),x,y) = [0 0 0]x + [1 0 0]y + [1] >= [0 0 0]x + [1 0 0]y + [0] = H(K1(x),K2(y))
                          [0 1 1]    [0 0 0]    [1]    [0 0 1]    [0 0 0]    [1]                 
   
             [1]    [0]      
   K1(A()) = [0] >= [0] = C()
             [0]    [0]      
   
             [0]    [0]      
   K2(B()) = [0] >= [0] = D()
             [1]    [0]      
  problem:
   K2(B()) -> D()
  Matrix Interpretation Processor:
   dimension: 3
   interpretation:
               [1 0 0]     [1]
    [K2](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [0],
    
          [0]
    [D] = [0]
          [0],
    
          [0]
    [B] = [0]
          [0]
   orientation:
              [1]    [0]      
    K2(B()) = [0] >= [0] = D()
              [0]    [0]      
   problem:
    
   Qed