YES

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

Proof:
 Extended Uncurrying Processor:
  application symbol: f
  symbol table:
   b ==> b0/0 b1/1
   c ==> c0/0 c1/1
   a ==> a0/0 a1/1
   
  uncurry-rules:
   f(a0(),x1) -> a1(x1)
   f(c0(),x3) -> c1(x3)
   f(b0(),x5) -> b1(x5)
  eta-rules:
   
  problem:
   a1(a1(x)) -> c1(b1(x))
   b1(b1(x)) -> a1(c1(x))
   c1(c1(x)) -> b1(a1(x))
   f(a0(),x1) -> a1(x1)
   f(c0(),x3) -> c1(x3)
   f(b0(),x5) -> b1(x5)
  Matrix Interpretation Processor: dim=3
   
   interpretation:
               [1 1 1]     [0]
    [b1](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [1],
    
           [0]
    [b0] = [1]
           [0],
    
               [1 0 1]     [0]
    [c1](x0) = [0 0 0]x0 + [0]
               [0 1 0]     [1],
    
           [0]
    [c0] = [0]
           [0],
    
               [1 1 1]     [0]
    [a1](x0) = [0 0 0]x0 + [0]
               [0 0 0]     [1],
    
           [0]
    [a0] = [0]
           [0],
    
                  [1 0 0]     [1 1 1]     [1]
    [f](x0, x1) = [0 1 0]x0 + [0 1 0]x1 + [0]
                  [0 0 0]     [0 1 0]     [1]
   orientation:
                [1 1 1]    [1]    [1 1 1]    [1]            
    a1(a1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = c1(b1(x))
                [0 0 0]    [1]    [0 0 0]    [1]            
    
                [1 1 1]    [1]    [1 1 1]    [1]            
    b1(b1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = a1(c1(x))
                [0 0 0]    [1]    [0 0 0]    [1]            
    
                [1 1 1]    [1]    [1 1 1]    [1]            
    c1(c1(x)) = [0 0 0]x + [0] >= [0 0 0]x + [0] = b1(a1(x))
                [0 0 0]    [1]    [0 0 0]    [1]            
    
                 [1 1 1]     [1]    [1 1 1]     [0]         
    f(a0(),x1) = [0 1 0]x1 + [0] >= [0 0 0]x1 + [0] = a1(x1)
                 [0 1 0]     [1]    [0 0 0]     [1]         
    
                 [1 1 1]     [1]    [1 0 1]     [0]         
    f(c0(),x3) = [0 1 0]x3 + [0] >= [0 0 0]x3 + [0] = c1(x3)
                 [0 1 0]     [1]    [0 1 0]     [1]         
    
                 [1 1 1]     [1]    [1 1 1]     [0]         
    f(b0(),x5) = [0 1 0]x5 + [1] >= [0 0 0]x5 + [0] = b1(x5)
                 [0 1 0]     [1]    [0 0 0]     [1]         
   problem:
    a1(a1(x)) -> c1(b1(x))
    b1(b1(x)) -> a1(c1(x))
    c1(c1(x)) -> b1(a1(x))
   DP Processor:
    DPs:
     a{1,#}(a1(x)) -> b{1,#}(x)
     a{1,#}(a1(x)) -> c{1,#}(b1(x))
     b{1,#}(b1(x)) -> c{1,#}(x)
     b{1,#}(b1(x)) -> a{1,#}(c1(x))
     c{1,#}(c1(x)) -> a{1,#}(x)
     c{1,#}(c1(x)) -> b{1,#}(a1(x))
    TRS:
     a1(a1(x)) -> c1(b1(x))
     b1(b1(x)) -> a1(c1(x))
     c1(c1(x)) -> b1(a1(x))
    TDG Processor:
     DPs:
      a{1,#}(a1(x)) -> b{1,#}(x)
      a{1,#}(a1(x)) -> c{1,#}(b1(x))
      b{1,#}(b1(x)) -> c{1,#}(x)
      b{1,#}(b1(x)) -> a{1,#}(c1(x))
      c{1,#}(c1(x)) -> a{1,#}(x)
      c{1,#}(c1(x)) -> b{1,#}(a1(x))
     TRS:
      a1(a1(x)) -> c1(b1(x))
      b1(b1(x)) -> a1(c1(x))
      c1(c1(x)) -> b1(a1(x))
     graph:
      c{1,#}(c1(x)) -> b{1,#}(a1(x)) ->
      b{1,#}(b1(x)) -> a{1,#}(c1(x))
      c{1,#}(c1(x)) -> b{1,#}(a1(x)) -> b{1,#}(b1(x)) -> c{1,#}(x)
      c{1,#}(c1(x)) -> a{1,#}(x) -> a{1,#}(a1(x)) -> c{1,#}(b1(x))
      c{1,#}(c1(x)) -> a{1,#}(x) -> a{1,#}(a1(x)) -> b{1,#}(x)
      b{1,#}(b1(x)) -> c{1,#}(x) -> c{1,#}(c1(x)) -> b{1,#}(a1(x))
      b{1,#}(b1(x)) -> c{1,#}(x) -> c{1,#}(c1(x)) -> a{1,#}(x)
      b{1,#}(b1(x)) -> a{1,#}(c1(x)) ->
      a{1,#}(a1(x)) -> c{1,#}(b1(x))
      b{1,#}(b1(x)) -> a{1,#}(c1(x)) -> a{1,#}(a1(x)) -> b{1,#}(x)
      a{1,#}(a1(x)) -> c{1,#}(b1(x)) ->
      c{1,#}(c1(x)) -> b{1,#}(a1(x))
      a{1,#}(a1(x)) -> c{1,#}(b1(x)) -> c{1,#}(c1(x)) -> a{1,#}(x)
      a{1,#}(a1(x)) -> b{1,#}(x) -> b{1,#}(b1(x)) -> a{1,#}(c1(x))
      a{1,#}(a1(x)) -> b{1,#}(x) -> b{1,#}(b1(x)) -> c{1,#}(x)
     Bounds Processor:
      bound: 0
      enrichment: match
      automaton:
       final states: {13,12,11,9,8,6,5,3,1}
       transitions:
        f130() -> 2*
        b{1,#,0}(10) -> 9*
        b{1,#,0}(2) -> 1*
        c{1,#,0}(2) -> 5*
        c{1,#,0}(4) -> 3*
        b{1,0}(10) -> 13*
        b{1,0}(2) -> 4*
        a{1,#,0}(7) -> 6*
        a{1,#,0}(2) -> 8*
        c{1,0}(2) -> 7*
        c{1,0}(4) -> 11*
        a{1,0}(7) -> 12*
        a{1,0}(2) -> 10*
        1 -> 8*
        3 -> 8*
        5 -> 1*
        6 -> 1*
        8 -> 5*
        9 -> 5*
        11 -> 10*
        12 -> 4*
        13 -> 7*
      problem:
       DPs:
        
       TRS:
        a1(a1(x)) -> c1(b1(x))
        b1(b1(x)) -> a1(c1(x))
        c1(c1(x)) -> b1(a1(x))
      Qed