YES

Problem:
 f(nil()) -> nil()
 f(.(nil(),y)) -> .(nil(),f(y))
 f(.(.(x,y),z)) -> f(.(x,.(y,z)))
 g(nil()) -> nil()
 g(.(x,nil())) -> .(g(x),nil())
 g(.(x,.(y,z))) -> g(.(.(x,y),z))

Proof:
 DP Processor:
  DPs:
   f#(.(nil(),y)) -> f#(y)
   f#(.(.(x,y),z)) -> f#(.(x,.(y,z)))
   g#(.(x,nil())) -> g#(x)
   g#(.(x,.(y,z))) -> g#(.(.(x,y),z))
  TRS:
   f(nil()) -> nil()
   f(.(nil(),y)) -> .(nil(),f(y))
   f(.(.(x,y),z)) -> f(.(x,.(y,z)))
   g(nil()) -> nil()
   g(.(x,nil())) -> .(g(x),nil())
   g(.(x,.(y,z))) -> g(.(.(x,y),z))
  Matrix Interpretation Processor: dim=3
   
   interpretation:
    [g#](x0) = [1 0 0]x0,
    
    [f#](x0) = [0 0 1]x0,
    
              [1 0 0]  
    [g](x0) = [0 1 0]x0
              [0 1 0]  ,
    
                  [1 0 0]     [0 1 0]     [0]
    [.](x0, x1) = [0 1 0]x0 + [0 1 0]x1 + [1]
                  [0 1 0]     [0 0 1]     [0],
    
              [1 0 1]  
    [f](x0) = [0 1 0]x0
              [0 1 0]  ,
    
            [0]
    [nil] = [1]
            [0]
   orientation:
    f#(.(nil(),y)) = [0 0 1]y + [1] >= [0 0 1]y = f#(y)
    
    f#(.(.(x,y),z)) = [0 1 0]x + [0 1 0]y + [0 0 1]z + [1] >= [0 1 0]x + [0 1 0]y + [0 0 1]z = f#(.(x,.(y,z)))
    
    g#(.(x,nil())) = [1 0 0]x + [1] >= [1 0 0]x = g#(x)
    
    g#(.(x,.(y,z))) = [1 0 0]x + [0 1 0]y + [0 1 0]z + [1] >= [1 0 0]x + [0 1 0]y + [0 1 0]z = g#(.(.(x,y),z))
    
               [0]    [0]        
    f(nil()) = [1] >= [1] = nil()
               [1]    [0]        
    
                    [0 1 1]    [1]    [0 1 0]    [0]                
    f(.(nil(),y)) = [0 1 0]y + [2] >= [0 1 0]y + [2] = .(nil(),f(y))
                    [0 1 0]    [2]    [0 1 0]    [1]                
    
                     [1 1 0]    [0 2 0]    [0 1 1]    [1]    [1 1 0]    [0 2 0]    [0 1 1]    [1]                 
    f(.(.(x,y),z)) = [0 1 0]x + [0 1 0]y + [0 1 0]z + [2] >= [0 1 0]x + [0 1 0]y + [0 1 0]z + [2] = f(.(x,.(y,z)))
                     [0 1 0]    [0 1 0]    [0 1 0]    [2]    [0 1 0]    [0 1 0]    [0 1 0]    [2]                 
    
               [0]    [0]        
    g(nil()) = [1] >= [1] = nil()
               [1]    [0]        
    
                    [1 0 0]    [1]    [1 0 0]    [1]                
    g(.(x,nil())) = [0 1 0]x + [2] >= [0 1 0]x + [2] = .(g(x),nil())
                    [0 1 0]    [2]    [0 1 0]    [0]                
    
                     [1 0 0]    [0 1 0]    [0 1 0]    [1]    [1 0 0]    [0 1 0]    [0 1 0]    [0]                 
    g(.(x,.(y,z))) = [0 1 0]x + [0 1 0]y + [0 1 0]z + [2] >= [0 1 0]x + [0 1 0]y + [0 1 0]z + [2] = g(.(.(x,y),z))
                     [0 1 0]    [0 1 0]    [0 1 0]    [2]    [0 1 0]    [0 1 0]    [0 1 0]    [2]                 
   problem:
    DPs:
     
    TRS:
     f(nil()) -> nil()
     f(.(nil(),y)) -> .(nil(),f(y))
     f(.(.(x,y),z)) -> f(.(x,.(y,z)))
     g(nil()) -> nil()
     g(.(x,nil())) -> .(g(x),nil())
     g(.(x,.(y,z))) -> g(.(.(x,y),z))
   Qed