YES

Problem:
 f(a(),b()) -> f(a(),c())
 f(c(),d()) -> f(b(),d())

Proof:
 Uncurry Processor:
  a1(b()) -> a1(c())
  c1(d()) -> b1(d())
  f(a(),x0) -> a1(x0)
  f(b(),x0) -> b1(x0)
  f(c(),x0) -> c1(x0)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
               [1 0 1]  
    [c1](x0) = [0 0 0]x0
               [0 0 0]  ,
    
               [1 0 0]  
    [b1](x0) = [0 0 0]x0
               [0 0 0]  ,
    
               [1 0 1]     [0]
    [a1](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [1],
    
          [0]
    [d] = [0]
          [1],
    
          [0]
    [c] = [1]
          [0],
    
                  [1 0 1]     [1 0 1]  
    [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                  [0 1 1]     [0 0 0]  ,
    
          [1]
    [b] = [0]
          [0],
    
          [0]
    [a] = [0]
          [1]
   orientation:
              [1]    [0]          
    a1(b()) = [0] >= [0] = a1(c())
              [1]    [1]          
    
              [1]    [0]          
    c1(d()) = [0] >= [0] = b1(d())
              [0]    [0]          
    
                [1 0 1]     [1]    [1 0 1]     [0]         
    f(a(),x0) = [0 0 0]x0 + [0] >= [0 0 0]x0 + [0] = a1(x0)
                [0 0 0]     [1]    [0 0 0]     [1]         
    
                [1 0 1]     [1]    [1 0 0]           
    f(b(),x0) = [0 0 0]x0 + [0] >= [0 0 0]x0 = b1(x0)
                [0 0 0]     [0]    [0 0 0]           
    
                [1 0 1]     [0]    [1 0 1]           
    f(c(),x0) = [0 0 0]x0 + [0] >= [0 0 0]x0 = c1(x0)
                [0 0 0]     [1]    [0 0 0]           
   problem:
    f(c(),x0) -> c1(x0)
   Matrix Interpretation Processor: dim=3
    
    interpretation:
                [1 0 0]  
     [c1](x0) = [0 0 0]x0
                [0 0 0]  ,
     
           [0]
     [c] = [1]
           [0],
     
                   [1 1 0]     [1 0 0]  
     [f](x0, x1) = [0 0 0]x0 + [0 0 0]x1
                   [0 0 0]     [0 0 0]  
    orientation:
                 [1 0 0]     [1]    [1 0 0]           
     f(c(),x0) = [0 0 0]x0 + [0] >= [0 0 0]x0 = c1(x0)
                 [0 0 0]     [0]    [0 0 0]           
    problem:
     
    Qed