Switches again

Now you should be worrying about whether this Boolean algebra has anything to do with switches at all. In the section on switches we saw that we had a `$+$' and a `$\times$', which were defined in terms of the parallel and series organisation of switches. Given that the 0 and 1 of the smallest Boolean algebra developed from switches, it seems natural to worry about whether switching circuits satisfy the axioms of a Boolean algebra. Let's work through them to see if any fail.

This is just the way that parallel and series switching works for $x$ and its complement.


  1. Show that for $x$ switch $x, x \times 1 = x$.
  2. Show that for switches $x, y$ and $z, x + yz = (x + y)(x + z)$.
  3. Show that for switches $x$ and $y, x + xy = x$ and $x(x + y) = x$.
  4. Show that the principle of duality holds.

We now have all the axioms of a Boolean algebra, so a set of switches joined in parallel and series forms a Boolean algebra. BUT, what use is this?

Let's consider an example. Recall Q3, Voting for Three of the 'Light in the Passage' section (see page 8). We have to set up switches so that a light goes on when two people out of three vote for an option by switching their switches on. Let $x$, $y$ and $z$ be the three options. Then $xy + xz + xz + xyz$ will give us a perfectly good switching system. However, we can simplify this using the properties of Boolean algebra.

$xy + xz + yz + xyz = x(y + z) + yz(1 + x) = x(y +z) + yz$.

This produces the switching system of Figure 5.

Switching circuit

Figure 5:A switching circuit for voting for three

For what it's worth, this switching set up requires one less switch than the number needed using the obvious alternative arrangement of $xy + xz + yz.$


  1. Use Boolean algebra to simplify the system of switches from Q1.
  2. Where possible, use Boolean algebra to simplify the switching circuits from Q10.
  3. Produce a switching system for the Passage problem (see Q11).

Next page - Content - Functions for Boolean Algebras