YES

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

Proof:
 DP Processor:
  DPs:
   a#(c(x1)) -> a#(x1)
   a#(c(x1)) -> b#(c(c(a(x1))))
   b#(c(x1)) -> b#(x1)
   b#(c(x1)) -> a#(b(x1))
  TRS:
   a(b(x1)) -> x1
   a(c(x1)) -> b(c(c(a(x1))))
   b(c(x1)) -> a(b(x1))
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b#](x0) = x0 + 3/2,
    
    [a#](x0) = 4x0,
    
    [c](x0) = 2x0 + 2,
    
    [a](x0) = 2x0,
    
    [b](x0) = 1/2x0
   orientation:
    a#(c(x1)) = 8x1 + 8 >= 4x1 = a#(x1)
    
    a#(c(x1)) = 8x1 + 8 >= 8x1 + 15/2 = b#(c(c(a(x1))))
    
    b#(c(x1)) = 2x1 + 7/2 >= x1 + 3/2 = b#(x1)
    
    b#(c(x1)) = 2x1 + 7/2 >= 2x1 = a#(b(x1))
    
    a(b(x1)) = x1 >= x1 = x1
    
    a(c(x1)) = 4x1 + 4 >= 4x1 + 3 = b(c(c(a(x1))))
    
    b(c(x1)) = x1 + 1 >= x1 = a(b(x1))
   problem:
    DPs:
     
    TRS:
     a(b(x1)) -> x1
     a(c(x1)) -> b(c(c(a(x1))))
     b(c(x1)) -> a(b(x1))
   Qed