YES

Problem:
 merge(x,nil()) -> x
 merge(nil(),y) -> y
 merge(++(x,y),++(u(),v())) -> ++(x,merge(y,++(u(),v())))
 merge(++(x,y),++(u(),v())) -> ++(u(),merge(++(x,y),v()))

Proof:
 DP Processor:
  DPs:
   merge#(++(x,y),++(u(),v())) -> merge#(y,++(u(),v()))
   merge#(++(x,y),++(u(),v())) -> merge#(++(x,y),v())
  TRS:
   merge(x,nil()) -> x
   merge(nil(),y) -> y
   merge(++(x,y),++(u(),v())) -> ++(x,merge(y,++(u(),v())))
   merge(++(x,y),++(u(),v())) -> ++(u(),merge(++(x,y),v()))
  Usable Rule Processor:
   DPs:
    merge#(++(x,y),++(u(),v())) -> merge#(y,++(u(),v()))
    merge#(++(x,y),++(u(),v())) -> merge#(++(x,y),v())
   TRS:
    
   Matrix Interpretation Processor: dim=4
    
    usable rules:
     
    interpretation:
     [merge#](x0, x1) = [1 1 1 1]x0 + [0 0 1 0]x1 + [1],
     
           [1]
           [1]
     [v] = [0]
           [1],
     
           [1]
           [1]
     [u] = [0]
           [0],
     
                    [1 1 0 0]     [1 0 1 0]     [0]
                    [0 0 0 0]     [0 1 1 1]     [1]
     [++](x0, x1) = [0 0 0 0]x0 + [1 0 0 0]x1 + [0]
                    [1 1 0 0]     [0 1 0 0]     [0]
    orientation:
     merge#(++(x,y),++(u(),v())) = [2 2 0 0]x + [2 2 2 1]y + [3] >= [1 1 1 1]y + [2] = merge#(y,++(u(),v()))
     
     merge#(++(x,y),++(u(),v())) = [2 2 0 0]x + [2 2 2 1]y + [3] >= [2 2 0 0]x + [2 2 2 1]y + [2] = merge#(++(x,y),v())
    problem:
     DPs:
      
     TRS:
      
    Qed