YES

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

Proof:
 AT confluence processor
  Complete TRS T' of input TRS:
  b(c(x)) -> a(x)
  b(b(x)) -> a(c(x))
  a(x) -> c(b(x))
  c(c(c(x))) -> b(x)
  
   T' = (P union S) with
  
   TRS P:
  
   TRS S:b(c(x)) -> a(x)
         b(b(x)) -> a(c(x))
         a(x) -> c(b(x))
         c(c(c(x))) -> b(x)
  
  S is linear and P is reversible.
  
   CP(S,S) = 
  b(a(x70)) = a(c(c(x70))), b(a(c(x71))) = a(c(b(x71))), b(b(x72)) = 
  a(c(c(x72))), c(b(x73)) = b(c(x73)), c(c(b(x74))) = b(c(c(x74)))
  
   CP(S,P union P^-1) = 
  
   CP(P union P^-1,S) = 
  
  
  We have to check termination of S:
  
  Matrix Interpretation Processor: dim=1
   
   interpretation:
    [b](x0) = x0 + 4,
    
    [c](x0) = x0 + 2,
    
    [a](x0) = x0 + 6
   orientation:
    b(c(x)) = x + 6 >= x + 6 = a(x)
    
    b(b(x)) = x + 8 >= x + 8 = a(c(x))
    
    a(x) = x + 6 >= x + 6 = c(b(x))
    
    c(c(c(x))) = x + 6 >= x + 4 = b(x)
   problem:
    b(c(x)) -> a(x)
    b(b(x)) -> a(c(x))
    a(x) -> c(b(x))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [b](x0) = 2x0 + 7,
     
     [c](x0) = x0,
     
     [a](x0) = 2x0 + 7
    orientation:
     b(c(x)) = 2x + 7 >= 2x + 7 = a(x)
     
     b(b(x)) = 4x + 21 >= 2x + 7 = a(c(x))
     
     a(x) = 2x + 7 >= 2x + 7 = c(b(x))
    problem:
     b(c(x)) -> a(x)
     a(x) -> c(b(x))
    Matrix Interpretation Processor: dim=1
     
     interpretation:
      [b](x0) = 2x0 + 1,
      
      [c](x0) = 2x0 + 2,
      
      [a](x0) = 4x0 + 5
     orientation:
      b(c(x)) = 4x + 5 >= 4x + 5 = a(x)
      
      a(x) = 4x + 5 >= 4x + 4 = c(b(x))
     problem:
      b(c(x)) -> a(x)
     Matrix Interpretation Processor: dim=1
      
      interpretation:
       [b](x0) = 2x0 + 1,
       
       [c](x0) = 2x0 + 1,
       
       [a](x0) = 4x0
      orientation:
       b(c(x)) = 4x + 3 >= 4x = a(x)
      problem:
       
      Qed