YES

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

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