What is the minimum number of NOR gates needed to implement a 2-input multiplexer?

In the first chapter of The Elements of Computing Systems, the reader is asked to implement basic logic gates like Not, And, Or, Xor, as well as a 2-input multiplexer and de-multiplexer. The authors instruct the reader to build their gates using the universal NAND gate as a basis.

I am able to build all of the requested gates using NAND gates without issue. Using truth tables and DeMorgan's law, I have minimized the gates as much as possible (e.g. Xor can be minimally represented using only 4 NAND gates, as can a 2-input Mux chip).

For the sake of rigor, I decided to build the gates using the other universal gate: NOR. In doing so, I noticed that NAND and NOR representations are pretty much tied in terms of complexity. For example, And gates require 2 NAND gates or 3 NOR gates, but this is countered by Or gates needing 2 NOR gates or 3 NAND gates.

For the 16 boolean functions that exist for 2 inputs, NAND and NOR end up tied. Neither gate is superior in terms of minimizing complexity. However, this doesn't seem to hold when building the Mux chip.

I can build the Mux chip in 4 NAND gates. The de-multiplexer (DMux) chip can be built in either 5 NAND gates or 4 NOR gates. My brain's pattern-recognition wants so very badly to think that the Mux chip must have a NOR representation in only 5 gates, thus making Mux and DMux another set of tied pairs leaving NAND and NOR gates tied once again.

But try as I may, I cannot find a way to implement a Mux chip in any fewer than 7 NOR gates.

Am I missing some minimization trick (as with masks in NAND-Xor or NOR-Xnor)?