YES

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

Proof:
 Matrix Interpretation Processor: dim=1
  
  interpretation:
   [c](x0) = 2x0 + 3,
   
   [b](x0) = x0 + 1,
   
   [a](x0) = 2x0 + 4
  orientation:
   b(a(a(x1))) = 4x1 + 13 >= 4x1 + 12 = a(b(c(x1)))
   
   c(a(x1)) = 4x1 + 11 >= 4x1 + 10 = a(c(x1))
   
   b(c(a(x1))) = 4x1 + 12 >= 4x1 + 12 = a(b(c(x1)))
   
   c(b(x1)) = 2x1 + 5 >= 2x1 + 5 = b(a(x1))
   
   a(c(b(x1))) = 4x1 + 14 >= 4x1 + 13 = c(b(a(x1)))
  problem:
   b(c(a(x1))) -> a(b(c(x1)))
   c(b(x1)) -> b(a(x1))
  Arctic Interpretation Processor:
   dimension: 2
   interpretation:
              [2  -&]  
    [c](x0) = [2  0 ]x0,
    
              [0 2]  
    [b](x0) = [0 0]x0,
    
              [0  2 ]  
    [a](x0) = [-& 0 ]x0
   orientation:
                  [4 6]      [4 2]                
    b(c(a(x1))) = [2 4]x1 >= [2 0]x1 = a(b(c(x1)))
    
               [2 4]      [0 2]             
    c(b(x1)) = [2 4]x1 >= [0 2]x1 = b(a(x1))
   problem:
    b(c(a(x1))) -> a(b(c(x1)))
   KBO Processor:
    weight function:
     w0 = 1
     w(c) = w(b) = w(a) = 1
    precedence:
     b > c ~ a
    problem:
     
    Qed