Can I simplify this to a 2-to-1 multiplexer?
Building L using a single 2-to-1 mux... and other logic

First, it is useful to construct the Karnaugh map based on your already constructed truth table:

S0 \ S1 S2
    | 00 01 11 10
 0  |  0  1  0  1
 1  |  1  0  1  0


Solving for the minimum sum of products form we get

\$ L = S_0 \overline{S_1} \overline{S_2} + S_0 S_1 S_2 + \overline{S_0} \overline{S_1} S_2 + \overline{S_0} S_1 \overline{S_2} \$

And note that the equation for a 2-to-1 mux would be

\$ y = x_1 x_2 + \overline{x_1} x_3 \$







simulate this circuit &ndash; Schematic created using CircuitLab

It is not possible to simplify L into the form for y. There is simply no way to do it, even if you invert any combination of the inputs. If you were allowed to use additional logic it would be possible. For example, we simplify 

\$ L = S_0 \overline{S_1} \overline{S_2} + S_0 S_1 S_2 + \overline{S_0} \overline{S_1} S_2 + \overline{S_0} S_1 \overline{S_2} \$

By factoring out the \$ S_0 \$ and \$\overline{S_0} \$ terms:

\$ L = S_0 (\overline{S_1} \overline{S_2} + S_1 S_2) + \overline{S_0} (\overline{S_1} S_2 + S_1 \overline{S_2}) \$

Simplifying the inner terms with the XOR and XNOR functions:

\$ L = S_0 (\overline{S_1 \oplus S_2}) + \overline{S_0} (S_1 \oplus S_2) \$

Which is pretty straightforward to build this circuit:







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This is the closest I could get to what I think you are asking for. 

An interesting simplification

Notice that the inner terms are complementary:

\$ L = S_0 (\overline{S_1 \oplus S_2}) + \overline{S_0} (S_1 \oplus S_2) \$

If we set the inner term as another variable, \$ A \$,

\$ A = S_1 \oplus S_2 \$

Then the equation for L becomes:

\$ L = S_0 \overline{A} + \overline{S_0} A \$

Fascinating. Do you see it? Its again the equation for XOR:

\$ L = S_0 \oplus A \$

Expanding \$ A \$ we get

\$ L = S_0 \oplus (S_1 \oplus S_2) \$

\$ L = S_0 \oplus S_1 \oplus S_2 \$







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