I'm looking at the proof of the PBW theorem for restricted Lie algebras to be found in Ponto and May's "More Concise Algebraic Topology", page 361 (367 in linked file). I either see an error in their proof, or fail to see my own error.
I believe that the counterexample is easy to set up: suppose you have a restricted Lie algebra $L$ over $F_2$ containing $x$ and $y$, such that $x$, $y$, $\xi(y)$ and $[x,\xi(y)]$ are linearly independent. In stating the PBW theorem one chooses an ordered basis of $L$ - let's make these four elements be basis elements, and order them $\xi(y)<x<y<[x,\xi(y)]$. Extend this ordered linearly independent set to an ordered basis of $L$ any way you like.
Right at the end of the proof, we are to see that $\tau\overline{\sigma}(I)=0$, but I do not think that this is so - isn't it true that $x(y^2+\xi(y))\in I$, and $$x(y^2+\xi(y))=xy^2+\xi(y)x+[x,\xi(y)]\overset{\tau\overline{\sigma}}{\mapsto}[x,\xi(y)],$$ which is nonzero in $B(L)$?
Perhaps, even if $\tau\overline{\sigma}(I)=0$ does fail, there's some clever way to produce a map in place of $E^0\overline{\tau}$ without ever producing $\overline{\tau}$, completing the proof, but I'm finding this rather difficult to do.