YES

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

Proof:
 DP Processor:
  DPs:
   b#(c(x1)) -> a#(x1)
   b#(b(x1)) -> c#(x1)
   b#(b(x1)) -> a#(c(x1))
   a#(x1) -> b#(x1)
   a#(x1) -> c#(b(x1))
   c#(c(c(x1))) -> b#(x1)
  TRS:
   b(c(x1)) -> a(x1)
   b(b(x1)) -> a(c(x1))
   a(x1) -> c(b(x1))
   c(c(c(x1))) -> b(x1)
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [c#](x0) = 2x0,
    
    [a#](x0) = 2x0 + 9/2,
    
    [b#](x0) = 2x0 + 3,
    
    [a](x0) = x0 + 3,
    
    [b](x0) = x0 + 2,
    
    [c](x0) = x0 + 1
   orientation:
    b#(c(x1)) = 2x1 + 5 >= 2x1 + 9/2 = a#(x1)
    
    b#(b(x1)) = 2x1 + 7 >= 2x1 = c#(x1)
    
    b#(b(x1)) = 2x1 + 7 >= 2x1 + 13/2 = a#(c(x1))
    
    a#(x1) = 2x1 + 9/2 >= 2x1 + 3 = b#(x1)
    
    a#(x1) = 2x1 + 9/2 >= 2x1 + 4 = c#(b(x1))
    
    c#(c(c(x1))) = 2x1 + 4 >= 2x1 + 3 = b#(x1)
    
    b(c(x1)) = x1 + 3 >= x1 + 3 = a(x1)
    
    b(b(x1)) = x1 + 4 >= x1 + 4 = a(c(x1))
    
    a(x1) = x1 + 3 >= x1 + 3 = c(b(x1))
    
    c(c(c(x1))) = x1 + 3 >= x1 + 2 = b(x1)
   problem:
    DPs:
     
    TRS:
     b(c(x1)) -> a(x1)
     b(b(x1)) -> a(c(x1))
     a(x1) -> c(b(x1))
     c(c(c(x1))) -> b(x1)
   Qed