YES

Problem:
 purge(nil()) -> nil()
 purge(.(x,y)) -> .(x,purge(remove(x,y)))
 remove(x,nil()) -> nil()
 remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))

Proof:
 DP Processor:
  DPs:
   purge#(.(x,y)) -> remove#(x,y)
   purge#(.(x,y)) -> purge#(remove(x,y))
   remove#(x,.(y,z)) -> remove#(x,z)
  TRS:
   purge(nil()) -> nil()
   purge(.(x,y)) -> .(x,purge(remove(x,y)))
   remove(x,nil()) -> nil()
   remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))
  Usable Rule Processor:
   DPs:
    purge#(.(x,y)) -> remove#(x,y)
    purge#(.(x,y)) -> purge#(remove(x,y))
    remove#(x,.(y,z)) -> remove#(x,z)
   TRS:
    remove(x,nil()) -> nil()
    remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))
   KBO Processor:
    argument filtering:
     pi(nil) = []
     pi(.) = [0,1]
     pi(remove) = [0,1]
     pi(=) = 0
     pi(if) = 0
     pi(purge#) = 0
     pi(remove#) = 1
    usable rules:
     remove(x,nil()) -> nil()
     remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))
    weight function:
     w0 = 1
     w(remove#) = w(purge#) = w(if) = w(=) = w(nil) = 1
     w(remove) = w(.) = 0
    precedence:
     remove# ~ purge# ~ if ~ . ~ nil > = ~ remove
    problem:
     DPs:
      
     TRS:
      remove(x,nil()) -> nil()
      remove(x,.(y,z)) -> if(=(x,y),remove(x,z),.(y,remove(x,z)))
    Qed