YES

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

Proof:
 DP Processor:
  DPs:
   f#(a(),b()) -> f#(a(),c())
   f#(c(),d()) -> f#(b(),d())
  TRS:
   f(a(),b()) -> f(a(),c())
   f(c(),d()) -> f(b(),d())
  Matrix Interpretation Processor: dim=4
   
   interpretation:
    [f#](x0, x1) = [0 1 0 0]x0 + [1 0 1 1]x1 + [1],
    
          [0]
          [0]
    [d] = [1]
          [0],
    
          [0]
          [1]
    [c] = [0]
          [1],
    
                  [0 0 1 0]     [1 1 0 1]     [0]
                  [0 0 0 0]     [1 1 0 0]     [0]
    [f](x0, x1) = [0 0 1 0]x0 + [0 0 1 1]x1 + [1]
                  [1 1 1 0]     [0 0 0 1]     [0],
    
          [1]
          [0]
    [b] = [0]
          [1],
    
          [0]
          [0]
    [a] = [1]
          [0]
   orientation:
    f#(a(),b()) = 3 >= 2 = f#(a(),c())
    
    f#(c(),d()) = 3 >= 2 = f#(b(),d())
    
                 [3]    [3]             
                 [1]    [1]             
    f(a(),b()) = [3] >= [3] = f(a(),c())
                 [2]    [2]             
    
                 [0]    [0]             
                 [0]    [0]             
    f(c(),d()) = [2] >= [2] = f(b(),d())
                 [1]    [1]             
   problem:
    DPs:
     
    TRS:
     f(a(),b()) -> f(a(),c())
     f(c(),d()) -> f(b(),d())
   Qed