YES

TRS:

  from(X) -> cons(X,n__from(s(X)))
  first(0(),Z) -> nil()
  first(s(X),cons(Y,Z)) -> cons(Y,n__first(X,activate(Z)))
  sel(0(),cons(X,Z)) -> X
  sel(s(X),cons(Y,Z)) -> sel(X,activate(Z))
  from(X) -> n__from(X)
  first(X1,X2) -> n__first(X1,X2)
  activate(n__from(X)) -> from(X)
  activate(n__first(X1,X2)) -> first(X1,X2)
  activate(X) -> X

max/plus interpretations on N:

  from_A(x1) = max{3, 4 + x1}
  from#_A(x1) = max{3, 4 + x1}
  cons_A(x1,x2) = max{4, x1, 2 + x2}
  cons#_A(x1,x2) = max{4, x1, 2 + x2}
  n__from_A(x1) = max{2, 2 + x1}
  n__from#_A(x1) = max{2, 2 + x1}
  s_A(x1) = max{0, x1}
  s#_A(x1) = max{0, x1}
  first_A(x1,x2) = max{6, 7 + x1, 2 + x2}
  first#_A(x1,x2) = max{6, 7 + x1, 2 + x2}
  0_A = 1
  0#_A = 1
  nil_A = 0
  nil#_A = 0
  n__first_A(x1,x2) = max{1, 5 + x1, x2}
  n__first#_A(x1,x2) = max{1, 5 + x1, x2}
  activate_A(x1) = max{5, 2 + x1}
  activate#_A(x1) = max{5, 2 + x1}
  sel_A(x1,x2) = max{11, 10 + x1, 6 + x2}
  sel#_A(x1,x2) = max{11, 10 + x1, 6 + x2}

precedence:

  0 = sel > n__from = nil = activate > from = first > cons = s = n__first