I have been studying quantum groups from Christian Kassel's Quantum Groups side-by-side with Majid's Foundations of Quantum Group Theory. I noticed that while Majid proves that the antipode of a quasi-triangular Hopf algebra is invertible, Kassel adds this as a condition to a few propositions. It seems as if Kassel takes the invertibility of the antipode as an independent assumption, which seems to subtly contradict Majid's treatment.
To provide context, allow me to repeat the relevant definitions and propositions. Both authors use standard definitions of Hopf algebras. Both authors define a quasi-triangular bialgebra or Hopf algebra $H$ as one such that there is an invertible element $R\in H\otimes H$ that satisfies $$\tau\circ \Delta x = R\Delta(x)R^{-1}\tag{1}\label{1}$$ for all $x\in H$ and $$ \begin{align} (\Delta\otimes\mathrm{id})R = R_{13}R_{23} && \text{and} && (\mathrm{id}\otimes\Delta)R = R_{13}R_{12}\ .\tag{2}\label{2} \end{align} $$ Kassel also calls bialgebras/Hopf algebras that only satisfy \eqref{1} quasi-cocommutative, and uses the term braided where Majid uses quasi-triangular, while acknowledging the latter usage in literature.
Later in Theorem VIII.2.4(b), Kassel explicitly includes the condition "If, moreover, (a braided bialgebra) $H$ has an invertible antipode". In the next section, right before Proposition VIII.3.1, he also comments that the last two equalities in equation (3.2) only hold when $H$ has an invertible antipode.
On the other hand, Majid proves (as his Proposition 2.1.8) that the antipode of a quasi-triangular Hopf algebra is invertible, using basic properties of the antipode and the counit, along with the axioms \eqref{1} and \eqref{2}, their consequences the Yang-Baxter equation and some additional facts proven in his Lemma 2.1.2.
Now the only real difference that I can see between the two treatments up to this point is that Kassel wants the property $(\mathrm{id}\otimes S^{-1})R = R^{-1}$ (Theorem VIII.2.4(b)) under the assumption that $S$ is invertible, whereas Majid wants $(\mathrm{id}\otimes S)R^{-1} = R$ without extra assumptions besides that $H$ be quasi-triangular (Lemma 2.1.2). Combined with the property $(S\otimes \mathrm{id})R = R^{-1}$, which both authors agree on, and which seemingly follows without using the assumption that $S$ is invertible in Kassel's case, we get an additional result $(S\otimes S)R = R$. This result is later used by Majid to prove that $S$ is invertible. So could it be that Majid's proof of $(\mathrm{id}\otimes S)R^{-1} = R$ doesn't actually follow unless you assume the invertibility of the antipode? Or is it possible that the invertibility assumption is actually redudant?
I am aware of this answer that also noticed this difference, but that answer and the discussion below don't really pin down which treatment makes more sense, either.
I have also checked the errata for Kassel's book, and it actually includes an entry that adds the condition "with invertible antipode" to a proposition involving braided Hopf algebras. This further shows that he doesn't consider this condition redundant.