YES

Problem:
 a(b(x1)) -> b(a(a(x1)))
 b(x1) -> c(a(c(x1)))
 a(a(x1)) -> a(c(a(x1)))

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