YES

Problem:
 del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
 f(true(),x,y,z) -> del(.(y,z))
 f(false(),x,y,z) -> .(x,del(.(y,z)))
 =(nil(),nil()) -> true()
 =(.(x,y),nil()) -> false()
 =(nil(),.(y,z)) -> false()
 =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))

Proof:
 DP Processor:
  DPs:
   del#(.(x,.(y,z))) -> =#(x,y)
   del#(.(x,.(y,z))) -> f#(=(x,y),x,y,z)
   f#(true(),x,y,z) -> del#(.(y,z))
   f#(false(),x,y,z) -> del#(.(y,z))
   =#(.(x,y),.(u(),v())) -> =#(y,v())
   =#(.(x,y),.(u(),v())) -> =#(x,u())
  TRS:
   del(.(x,.(y,z))) -> f(=(x,y),x,y,z)
   f(true(),x,y,z) -> del(.(y,z))
   f(false(),x,y,z) -> .(x,del(.(y,z)))
   =(nil(),nil()) -> true()
   =(.(x,y),nil()) -> false()
   =(nil(),.(y,z)) -> false()
   =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))
  Usable Rule Processor:
   DPs:
    del#(.(x,.(y,z))) -> =#(x,y)
    del#(.(x,.(y,z))) -> f#(=(x,y),x,y,z)
    f#(true(),x,y,z) -> del#(.(y,z))
    f#(false(),x,y,z) -> del#(.(y,z))
    =#(.(x,y),.(u(),v())) -> =#(y,v())
    =#(.(x,y),.(u(),v())) -> =#(x,u())
   TRS:
    =(nil(),nil()) -> true()
    =(.(x,y),nil()) -> false()
    =(nil(),.(y,z)) -> false()
    =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))
   Arctic Interpretation Processor:
    dimension: 1
    usable rules:
     =(nil(),nil()) -> true()
     =(.(x,y),nil()) -> false()
     =(nil(),.(y,z)) -> false()
     =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))
    interpretation:
     [f#](x0, x1, x2, x3) = x0 + 2x1 + 5x2 + 3x3,
     
     [=#](x0, x1) = x1,
     
     [del#](x0) = x0,
     
     [and](x0, x1) = x0 + 0,
     
     [v] = 0,
     
     [u] = 1,
     
     [nil] = 4,
     
     [false] = 4,
     
     [true] = 0,
     
     [=](x0, x1) = 1x0 + 1x1,
     
     [.](x0, x1) = 4x0 + 2x1
    orientation:
     del#(.(x,.(y,z))) = 4x + 6y + 4z >= y = =#(x,y)
     
     del#(.(x,.(y,z))) = 4x + 6y + 4z >= 2x + 5y + 3z = f#(=(x,y),x,y,z)
     
     f#(true(),x,y,z) = 2x + 5y + 3z + 0 >= 4y + 2z = del#(.(y,z))
     
     f#(false(),x,y,z) = 2x + 5y + 3z + 4 >= 4y + 2z = del#(.(y,z))
     
     =#(.(x,y),.(u(),v())) = 5 >= 0 = =#(y,v())
     
     =#(.(x,y),.(u(),v())) = 5 >= 1 = =#(x,u())
     
     =(nil(),nil()) = 5 >= 0 = true()
     
     =(.(x,y),nil()) = 5x + 3y + 5 >= 4 = false()
     
     =(nil(),.(y,z)) = 5y + 3z + 5 >= 4 = false()
     
     =(.(x,y),.(u(),v())) = 5x + 3y + 6 >= 1x + 2 = and(=(x,u()),=(y,v()))
    problem:
     DPs:
      
     TRS:
      =(nil(),nil()) -> true()
      =(.(x,y),nil()) -> false()
      =(nil(),.(y,z)) -> false()
      =(.(x,y),.(u(),v())) -> and(=(x,u()),=(y,v()))
    Qed