 Solution

The printed input $x^r{x}^s$ is a genuine typo in the PDF as well as in the TeX: $E$ annihilates every power of $x$, whereas the displayed right-hand side is generally nonzero. The intended input is $x^r{y}^s$, and the calculation below proves that corrected identity.

The given comultiplications make $C_q$ a coalgebra, with counit one on $K,K^{-1},I$ and zero on $E,F$. Write $I_0=p(I)$, $K_0=p(K)$ and $E_0=p(E)$, and similarly for $F$. The measuring coalgebra identities areThe group-like elements act multiplicatively. Although the statement omits an explicit specification of $p(I)$, its identity action follows from the other data and the quantum plane relation. Applying $E_0$ to $yx=qxy$ gives $q{x}^2=q{I}_0(x)x$, so cancellation in the noncommutative domain gives $I_0(x)=x$. Applying $F_0$ to the same relation gives $q{y}^2=qy{I}_0(y)$, so $I_0(y)=y$. Thus $I_0$ is the identity on the entire quantum plane.

Now $E_0(x^r)=0$ by induction. Suppose $E_0(y^{s-1})=[s-1]x{y}^{s-2}$. The measuring rule and $y^{s-1}x=q^{s-1}x{y}^{s-1}$ giveThe induction starts with $E_0(y)=x$ and $[1]=1$. A final product rule givesFor $s=0$ the answer is zero. By contrast the literal printed input has $\boxed{p(E)(x^r{x}^s)=0}$; it cannot satisfy the printed right-hand side for $s=1$.