YES

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

Proof:
 DP Processor:
  DPs:
   a#(b(c(x1))) -> a#(b(x1))
   a#(b(c(x1))) -> a#(a(b(x1)))
  TRS:
   a(b(x1)) -> x1
   a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
  Arctic Interpretation Processor:
   dimension: 1
   usable rules:
    a(b(x1)) -> x1
    a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
   interpretation:
    [a](x0) = 3x0 + 4,
    
    [a#](x0) = x0 + 2,
    
    [b](x0) = -3x0 + 0,
    
    [c](x0) = 3x0 + 7
   orientation:
    a#(b(c(x1))) = x1 + 4 >= -3x1 + 2 = a#(b(x1))
    
    a#(b(c(x1))) = x1 + 4 >= x1 + 4 = a#(a(b(x1)))
    
    a(b(x1)) = x1 + 4 >= x1 = x1
    
    a(b(c(x1))) = 3x1 + 7 >= 3x1 + 7 = b(c(b(c(a(a(b(x1)))))))
   problem:
    DPs:
     a#(b(c(x1))) -> a#(a(b(x1)))
    TRS:
     a(b(x1)) -> x1
     a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
   Restore Modifier:
    DPs:
     a#(b(c(x1))) -> a#(a(b(x1)))
    TRS:
     a(b(x1)) -> x1
     a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
    Matrix Interpretation Processor: dim=1
     
     usable rules:
      a(b(x1)) -> x1
      a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
     interpretation:
      [a](x0) = 2x0,
      
      [a#](x0) = 9/2x0,
      
      [b](x0) = 1/2x0,
      
      [c](x0) = 2x0 + 1/2
     orientation:
      a#(b(c(x1))) = 9/2x1 + 9/8 >= 9/2x1 = a#(a(b(x1)))
      
      a(b(x1)) = x1 >= x1 = x1
      
      a(b(c(x1))) = 2x1 + 1/2 >= 2x1 + 1/2 = b(c(b(c(a(a(b(x1)))))))
     problem:
      DPs:
       
      TRS:
       a(b(x1)) -> x1
       a(b(c(x1))) -> b(c(b(c(a(a(b(x1)))))))
     Qed