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
   interpretation:
    [a#](x0) = x0 + 2,
    
    [c](x0) = 7x0 + 15,
    
    [a](x0) = 7x0 + 0,
    
    [b](x0) = -7x0 + 0
   orientation:
    a#(b(c(x1))) = x1 + 8 >= -7x1 + 2 = a#(b(x1))
    
    a#(b(c(x1))) = x1 + 8 >= x1 + 7 = a#(a(b(x1)))
    
    a(b(x1)) = x1 + 7 >= x1 = x1
    
    a(b(c(x1))) = 7x1 + 15 >= 7x1 + 14 = 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)))))))
   Matrix Interpretation Processor: dim=1
    
    interpretation:
     [a#](x0) = 4x0 + 2,
     
     [c](x0) = 2x0 + 1/2,
     
     [a](x0) = 2x0,
     
     [b](x0) = 1/2x0
    orientation:
     a#(b(c(x1))) = 4x1 + 3 >= 4x1 + 2 = 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