YES

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

Proof:
 Matrix Interpretation Processor: dim=3
  
  interpretation:
             [1 1 0]     [0]
   [a](x0) = [0 0 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]     [0]
   [b](x0) = [0 1 0]x0 + [1]
             [0 0 0]     [0],
   
             [1 0 0]  
   [c](x0) = [0 0 0]x0
             [0 0 0]  
  orientation:
              [1 1 0]     [1]    [1 1 0]     [0]              
   a(b(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 + [1] = b(c(a(x1)))
              [0 0 0]     [0]    [0 0 0]     [0]              
   
              [1 0 0]     [0]    [1 0 0]                
   b(c(x1)) = [0 0 0]x1 + [1] >= [0 0 0]x1 = c(b(b(x1)))
              [0 0 0]     [0]    [0 0 0]                
   
              [1 1 0]     [0]    [1 0 0]     [0]              
   b(a(x1)) = [0 0 0]x1 + [2] >= [0 0 0]x1 + [1] = a(c(b(x1)))
              [0 0 0]     [0]    [0 0 0]     [0]              
  problem:
   b(c(x1)) -> c(b(b(x1)))
   b(a(x1)) -> a(c(b(x1)))
  String Reversal Processor:
   c(b(x1)) -> b(b(c(x1)))
   a(b(x1)) -> b(c(a(x1)))
   WPO Processor:
    algebra: Sum
    weight function:
     w0 = 0
     w(c) = w(a) = w(b) = 0
    status function:
     st(c) = st(a) = st(b) = [0]
    precedence:
     a > c > b
    problem:
     
    Qed