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