YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   d ==> d0/0
   c ==> c0/0 c1/1
   b ==> b0/0 b1/1
   a ==> a0/0 a1/1
   
  uncurry-rules:
   f(a0(),x0) -> a1(x0)
   f(b0(),x2) -> b1(x2)
   f(c0(),x4) -> c1(x4)
  eta-rules:
   
  problem:
   a1(b0()) -> a1(c0())
   c1(d0()) -> b1(d0())
   f(a0(),x0) -> a1(x0)
   f(b0(),x2) -> b1(x2)
   f(c0(),x4) -> c1(x4)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
               [1 0 0]  
    [b1](x0) = [0 0 1]x0
               [0 0 0]  ,
    
           [0]
    [a0] = [0]
           [0],
    
               [1 1 1]  
    [c1](x0) = [0 0 1]x0
               [1 0 0]  ,
    
           [0]
    [b0] = [1]
           [0],
    
           [0]
    [d0] = [0]
           [1],
    
           [0]
    [c0] = [0]
           [0],
    
               [1 1 0]  
    [a1](x0) = [0 0 0]x0
               [0 0 0]  ,
    
                  [1 0 0]     [1 1 1]     [1]
    [f](x0, x1) = [0 0 0]x0 + [0 0 1]x1 + [0]
                  [0 0 0]     [1 0 0]     [0]
   orientation:
               [1]    [0]           
    a1(b0()) = [0] >= [0] = a1(c0())
               [0]    [0]           
    
               [1]    [0]           
    c1(d0()) = [1] >= [1] = b1(d0())
               [0]    [0]           
    
                 [1 1 1]     [1]    [1 1 0]           
    f(a0(),x0) = [0 0 1]x0 + [0] >= [0 0 0]x0 = a1(x0)
                 [1 0 0]     [0]    [0 0 0]           
    
                 [1 1 1]     [1]    [1 0 0]           
    f(b0(),x2) = [0 0 1]x2 + [0] >= [0 0 1]x2 = b1(x2)
                 [1 0 0]     [0]    [0 0 0]           
    
                 [1 1 1]     [1]    [1 1 1]           
    f(c0(),x4) = [0 0 1]x4 + [0] >= [0 0 1]x4 = c1(x4)
                 [1 0 0]     [0]    [1 0 0]           
   problem:
    
   Qed