Is a bialgebra pairing of Hopf algebras automatically a Hopf pairing?

The following question came up in the course on Quantum Groups here at UC Berkeley. (If you care, I have been TeXing uneditted lecture notes.)

Let $X,Y$ be (infinite-dimensional) Hopf algebras over a ground field $mathbb F$. A linear map $langle,rangle : Xotimes Y to mathbb F$ is a bialgebra pairing if $langle x,y_1y_2 rangle = langle Delta x,y_1otimes y_2rangle$ and $langle x_1x_2,yrangle = langle x_1otimes x_2,Delta yrangle$ for all $x,x_1,x_2 in X$ and $y,y_1,y_2 in Y$. (You must pick a convention of how to define the pairing $langle,rangle : X^{otimes 2} otimes Y^{otimes 2} to mathbb F$.) And we also demand that $langle 1,- rangle = epsilon_Y$ and $langle -,1rangle = epsilon_X$, but this might follow from the previous conditions. (See edit.)

A bialgebra pairing is Hopf if it also respects the antipode: $langle S(x),y rangle = langle x,S(y)rangle$. A pairing $langle,rangle : Xotimes Y to mathbb F$ is nondegenerate if each of the the induced maps $X to Y^*$ and $Y to X^*$ has trivial kernel.

Question: Is a (nondegenerate) bialgebra pairing of Hopf algebras necessarily Hopf? (Does it depend on whether the pairing is nondegenerate?)

My intuition is that regardless of the nondegeneracy, the answer is "Yes": my motivation is that a bialgebra homomorphism between Hopf algebras automatically respects the antipode. But we were unable to make this into a proof in the infinite-dimensional case.

Edit: If $langle,rangle: Xotimes Y to mathbb F$ is nondegenerate, then it is true that as soon as it satisfies $langle x,y_1y_2 rangle = langle Delta x,y_1otimes y_2rangle$ and $langle x_1x_2,yrangle = langle x_1otimes x_2,Delta yrangle$, so that the induced maps $X to Y^*$ and $Y to X^*$ are (possibly non-unital) algebra homomorphisms, then it also satisfies $langle 1,- rangle = epsilon_Y$ and $langle -,1rangle = epsilon_X$, so that the algebra homomorphism are actually unital. But I think that this does require that the pairing be nondegenerate. At least, I don't see how to prove it without the nondegeneracy assumption. So probably the nondegeneracy is required for the statement about antipodes as well.