YES

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

Proof:
 String Reversal Processor:
  b(a(a(x1))) -> a(a(a(b(c(x1)))))
  c(a(x1)) -> a(b(x1))
  DP Processor:
   DPs:
    b#(a(a(x1))) -> c#(x1)
    b#(a(a(x1))) -> b#(c(x1))
    c#(a(x1)) -> b#(x1)
   TRS:
    b(a(a(x1))) -> a(a(a(b(c(x1)))))
    c(a(x1)) -> a(b(x1))
   TDG Processor:
    DPs:
     b#(a(a(x1))) -> c#(x1)
     b#(a(a(x1))) -> b#(c(x1))
     c#(a(x1)) -> b#(x1)
    TRS:
     b(a(a(x1))) -> a(a(a(b(c(x1)))))
     c(a(x1)) -> a(b(x1))
    graph:
     c#(a(x1)) -> b#(x1) -> b#(a(a(x1))) -> b#(c(x1))
     c#(a(x1)) -> b#(x1) -> b#(a(a(x1))) -> c#(x1)
     b#(a(a(x1))) -> c#(x1) -> c#(a(x1)) -> b#(x1)
     b#(a(a(x1))) -> b#(c(x1)) -> b#(a(a(x1))) -> b#(c(x1))
     b#(a(a(x1))) -> b#(c(x1)) -> b#(a(a(x1))) -> c#(x1)
    Matrix Interpretation Processor: dim=4
     
     interpretation:
      [c#](x0) = [0 0 0 1]x0,
      
                [0 0 0 1]     [1]
                [1 0 0 0]     [0]
      [a](x0) = [0 0 0 0]x0 + [0]
                [0 1 0 0]     [0],
      
      [b#](x0) = [0 1 0 0]x0,
      
                [0 1 0 0]  
                [0 1 0 0]  
      [b](x0) = [0 0 0 1]x0
                [0 1 0 0]  ,
      
                [1 0 0 1]  
                [0 0 0 1]  
      [c](x0) = [0 0 0 0]x0
                [1 0 0 1]  
     orientation:
      b#(a(a(x1))) = [0 0 0 1]x1 + [1] >= [0 0 0 1]x1 = c#(x1)
      
      b#(a(a(x1))) = [0 0 0 1]x1 + [1] >= [0 0 0 1]x1 = b#(c(x1))
      
      c#(a(x1)) = [0 1 0 0]x1 >= [0 1 0 0]x1 = b#(x1)
      
                    [0 0 0 1]     [1]    [0 0 0 1]     [1]                    
                    [0 0 0 1]     [1]    [0 0 0 1]     [1]                    
      b(a(a(x1))) = [1 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(a(a(b(c(x1)))))
                    [0 0 0 1]     [1]    [0 0 0 1]     [1]                    
      
                 [0 1 0 1]     [1]    [0 1 0 0]     [1]           
                 [0 1 0 0]     [0]    [0 1 0 0]     [0]           
      c(a(x1)) = [0 0 0 0]x1 + [0] >= [0 0 0 0]x1 + [0] = a(b(x1))
                 [0 1 0 1]     [1]    [0 1 0 0]     [0]           
     problem:
      DPs:
       c#(a(x1)) -> b#(x1)
      TRS:
       b(a(a(x1))) -> a(a(a(b(c(x1)))))
       c(a(x1)) -> a(b(x1))
     Restore Modifier:
      DPs:
       c#(a(x1)) -> b#(x1)
      TRS:
       b(a(a(x1))) -> a(a(a(b(c(x1)))))
       c(a(x1)) -> a(b(x1))
      EDG Processor:
       DPs:
        c#(a(x1)) -> b#(x1)
       TRS:
        b(a(a(x1))) -> a(a(a(b(c(x1)))))
        c(a(x1)) -> a(b(x1))
       graph:
        
       SCC Processor:
        #sccs: 0
        #rules: 0
        #arcs: 0/1