Missing parentheses in $s(k (s I I))(s(\lambda y. s(k y))(\lambda y. s I I)$ leads to interesting error in an nLab page. Need a double check. I think I found an error in the nLab page on partial combinatory algebra in the Example combinators section:

Finally, consider the classical construction of the fixed-point combinator, $Y = \lambda y. (\lambda x. y(x x))(\lambda x. y(x x))$. We have firstly
$$\lambda x. y(x x) = s(\lambda x. y)(\lambda x. x x) = s(k y)(s I I)$$
which means
$$\array{
Y & = & \lambda y. (s I I)(s(k y)(s I I)) & (1)\\
 & = & s(\lambda y. s I I)(\lambda y. s(k y)(s I I)) & (2)\\ 
 & = & s(k (s I I))(s(\lambda y. s(k y))(\lambda y. s I I) & (3) \\ 
 & = & s(k (s I I))(s(s (k s)(\lambda y. k y)))(k (s I I)) & (4) \\ 
 & = & s(k (s I I))(s(s (k s)(s(k k)I)))(k (s I I)) & (5)
}$$

I think I found an error.
$(3)$ has a missing closing parenthesis, i.e. $\big)$. The closing parenthesis can go in one of two positions. The error stems from there.
$(2)$ is equivalent to $(3)$ with a trailing closing parenthesis (emphasized):
$$s(k (s I I))(s(\lambda y. s(k y))(\lambda y. s I I)\Big)\qquad(3a)$$
On the other hand, $(4)$ is equivalent to $(3)$ with a closing parenthesis in the other possible position (emphasized):
$$s(k (s I I))(s(\lambda y. s(k y))\Big)(\lambda y. s I I)\qquad (3b)$$

Can someone with patience confirm? I’d use the SKI combinator interpreter to double check, but it supports purely SKI input only (can’t mix in $\lambda$, variables, nor terms).
 A: I totally agree that what you have noticed is the only error among the equivalences, well spotted!
A missing parenthesis leads to the completely wrong conclusion that $S(K(SII))(S(S(KS)(S(KK)I)))(K(SII))$ is a fixed point combinator!
A proof that $S(K(SII))(S(S(KS)(S(KK)I)))(K(SII))$ is not a fixed point combinator is here.

To @mohottnad: Equivalence $(1)$ is$-$quite surprisingly$-$correct! Indeed,
$$Y = \lambda y. (\lambda x. y(xx)) (\lambda x. y(xx))$$
and we have that
\begin{align}
\lambda x. y(xx) &=_\beta S(Ky)(SII)
\\
\lambda x.xx &=_\beta SII
\end{align}
Since $(\lambda x.xx)(\lambda x.y(xx)) =_\beta (\lambda x.y(xx))(\lambda x.y(xx))$, we can conclude that
\begin{align}
Y &=_\beta \lambda y.(\lambda x.xx)(\lambda x.y(xx)) 
\\
&=_\beta \lambda y.(SII)(S(Ky)(SII))
\end{align}
I agree that the explanation on nLab's page could be a bit more lengthy and thorough. At first sight, I was convinced that equivalence $(1)$ was wrong, too.

An emended version of the equivalences in nLab's page about partial combinatory algebra, Point 4 in section Examples of Combinators, would be the following:
\begin{align}
Y & =_\beta \lambda y. (S I I)(S(K y)(S I I)) 
\\
 & =_\beta S(\lambda y. S I I)(\lambda y. S(K y)(S I I)) 
\\ 
 & =_\beta S(K(S I I)) \big(S(\lambda y. S(K y))(\lambda y. S I I)\big)
 \\ 
&=_\beta S(K(SII)) \big(S(S(KS)K)(K(SII))\big) 
\end{align}
where, in the last equivalence, we used the fact that
\begin{align}
S(S(KS)K)(K(SII))
&\to_\beta^* S (\lambda y.(KSy)(Ky)) (K(SII))
\\
&\to_\beta^*  S (\lambda y. S(Ky)) (K(SII))
\\
&\to_\beta S(\lambda y. S(K y))(\lambda y. S I I)
\end{align}
Therefore, $S(K(SII)) \big(S(S(KS)K)(K(SII))\big) $ is a fixed point combinator, and this is consistent with Wikipedia's page about fixed point combinators.
