A singlet-pairing superconductor is always also a super-spin-current-conductor.

A heuristic argument and a simple theory are used to show that, as a fundamental difference between BEC and BCS condensation of fermion pairs, \textit{the later, even for singlet pairing, can carry a sizable dissipation-less spin-current below practically the same T}$_{C}$. The heuristic argument is based on the similarity between a spin-current carried by a singlet-pairing condensate and (coherent) partner changing in a dancing hall. Simple theory: We consider singlet pairing in a normal metal carrying a moderate spin-current, which causes the spin-up- and -down Fermi surfaces (FSs) to be shifted in the momentum space by $\pm $\textbf{q}/2. [(k,$\uparrow )$,(-k,$\downarrow )$]-pairing is clearly still possible over the entire FSs. To favor a spin current in the system, we introduce a vector Lagrange multiplier \textbf{v}$_{sp}$, and add -\textbf{v}$_{sp}$\textbf{$\cdot \Sigma $ }$_{k, \sigma }\sigma $ h\textbf{k c}$_{ k,\sigma }^{\dag }$\textbf{c}$_{ k, \sigma }$to the Hamiltonian. Since time-reversal invariance is not broken, negligible changes to all properties of the singlet-pairing state follow, and the system remains fully gapped. No depairing can be induced even for a sizable spin current. Two experimental tests of this prediction will be discussed.