YES

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

Proof:
 Arctic Interpretation Processor:
  dimension: 2
  interpretation:
             [0  0 ]  
   [c](x0) = [-& 0 ]x0,
   
             [1  1 ]  
   [b](x0) = [-& 1 ]x0,
   
             [1  1 ]  
   [a](x0) = [-& 1 ]x0
  orientation:
           [1  1 ]      [1  1 ]          
   a(x1) = [-& 1 ]x1 >= [-& 1 ]x1 = b(x1)
   
              [2  2 ]      [2  2 ]                
   a(b(x1)) = [-& 2 ]x1 >= [-& 2 ]x1 = b(c(a(x1)))
   
           [1  1 ]      [0  0 ]          
   b(x1) = [-& 1 ]x1 >= [-& 0 ]x1 = c(x1)
   
              [1  1 ]      [1  1 ]          
   c(b(x1)) = [-& 1 ]x1 >= [-& 1 ]x1 = a(x1)
  problem:
   a(x1) -> b(x1)
   a(b(x1)) -> b(c(a(x1)))
   c(b(x1)) -> a(x1)
  KBO Processor:
   weight function:
    w0 = 1
    w(b) = w(a) = 1
    w(c) = 0
   precedence:
    c > a > b
   problem:
    
   Qed