YES

Problem:
 msort(nil()) -> nil()
 msort(.(x,y)) -> .(min(x,y),msort(del(min(x,y),.(x,y))))
 min(x,nil()) -> x
 min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
 del(x,nil()) -> nil()
 del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))

Proof:
 DP Processor:
  DPs:
   msort#(.(x,y)) -> del#(min(x,y),.(x,y))
   msort#(.(x,y)) -> msort#(del(min(x,y),.(x,y)))
   msort#(.(x,y)) -> min#(x,y)
   min#(x,.(y,z)) -> min#(y,z)
   min#(x,.(y,z)) -> min#(x,z)
   del#(x,.(y,z)) -> del#(x,z)
  TRS:
   msort(nil()) -> nil()
   msort(.(x,y)) -> .(min(x,y),msort(del(min(x,y),.(x,y))))
   min(x,nil()) -> x
   min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
   del(x,nil()) -> nil()
   del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))
  Usable Rule Processor:
   DPs:
    msort#(.(x,y)) -> del#(min(x,y),.(x,y))
    msort#(.(x,y)) -> msort#(del(min(x,y),.(x,y)))
    msort#(.(x,y)) -> min#(x,y)
    min#(x,.(y,z)) -> min#(y,z)
    min#(x,.(y,z)) -> min#(x,z)
    del#(x,.(y,z)) -> del#(x,z)
   TRS:
    min(x,nil()) -> x
    min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
    del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))
    del(x,nil()) -> nil()
   Matrix Interpretation Processor: dim=1
    
    usable rules:
     del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))
     del(x,nil()) -> nil()
    interpretation:
     [min#](x0, x1) = x1 + 2,
     
     [del#](x0, x1) = 2x1,
     
     [msort#](x0) = 2x0 + 7/2,
     
     [=](x0, x1) = 0,
     
     [if](x0, x1, x2) = 2x0 + 5/2,
     
     [<=](x0, x1) = x0 + 2x1,
     
     [del](x0, x1) = 5/2,
     
     [min](x0, x1) = x1 + 2,
     
     [.](x0, x1) = 2x1 + 3,
     
     [nil] = 1/2
    orientation:
     msort#(.(x,y)) = 4y + 19/2 >= 4y + 6 = del#(min(x,y),.(x,y))
     
     msort#(.(x,y)) = 4y + 19/2 >= 17/2 = msort#(del(min(x,y),.(x,y)))
     
     msort#(.(x,y)) = 4y + 19/2 >= y + 2 = min#(x,y)
     
     min#(x,.(y,z)) = 2z + 5 >= z + 2 = min#(y,z)
     
     min#(x,.(y,z)) = 2z + 5 >= z + 2 = min#(x,z)
     
     del#(x,.(y,z)) = 4z + 6 >= 2z = del#(x,z)
     
     min(x,nil()) = 5/2 >= x = x
     
     min(x,.(y,z)) = 2z + 5 >= 2x + 4y + 5/2 = if(<=(x,y),min(x,z),min(y,z))
     
     del(x,.(y,z)) = 5/2 >= 5/2 = if(=(x,y),z,.(y,del(x,z)))
     
     del(x,nil()) = 5/2 >= 1/2 = nil()
    problem:
     DPs:
      
     TRS:
      min(x,nil()) -> x
      min(x,.(y,z)) -> if(<=(x,y),min(x,z),min(y,z))
      del(x,.(y,z)) -> if(=(x,y),z,.(y,del(x,z)))
      del(x,nil()) -> nil()
    Qed