Boolean Algebra is related to digital electronics in several different ways. For example, consider this circuit.

First look at this circuit's truth table.

         _   _
 A B C | A | A * B | A * C | D
-------------------------------
 0 0 0 | 1 |   0   |   0   | 0
 0 0 1 | 1 |   0   |   0   | 0
 0 1 0 | 1 |   1   |   0   | 1
 0 1 1 | 1 |   1   |   0   | 1
 1 0 0 | 0 |   0   |   0   | 1
 1 0 1 | 0 |   0   |   1   | 1
 1 1 0 | 0 |   0   |   0   | 1
 1 1 1 | 0 |   0   |   1   | 1

Looking at the truth table, notice that D is true only when A or B is true. The first two lines of the truth table D is 0. But as soon as B becomes true, the rest of the table is true. When B isn't true anymore, A is and it's still a true output. Therefore, the Boolean Algebra formula is probably D=A+B.

The N-FETs of the transistor level schematic can also be helpful.  Remember that the condition that must be met is the same for a NOR as it is an OR, only the output changes.  Therefore, for our test, let's assume it is an inverted output.

From a current flow point of view, A is in two different paths. When A and C are both on (the middle path), path A was already on. It doesn't really matter what C is, since it requires a signal that ends up shorting that path out. As for the right path, when A is off then inverted A is on. Then the only requirement is B. Therefore, D is connected to ground if A is on, or if B is on. Again, the formula is probably D=A+B.

And finally, the Boolean Algebra manipulation.

          _
 D = A + (A * B) + (A * C)           [given formula]
      _
 D = (A * B) + A + (A * C)           [A + B = B + A]
      _
 D = (A * B) + (A * (1 + C))
                   [A * (B + C) = (A * B) + (A * C)]
      _
 D = (A * B) + (A * 1)                   [1 + A = 1]
      _
 D = (A * B) + A                         [1 * A = A]
          _
 D = A + (A * B)                     [A + B = B + A]
          _
 D = (A + A) * (A + B)
                   [A + (B * C) = (A + B) * (A + C)]

 D = 1 * (A + B)                         [A + A = 1]

 D = A + B                               [1 * A = A]

From investigating the circuit's truth table, and the transistor schematics, it is possible to figure out the formula. Using the Boolean Algebra rules, it is possible to prove the investigation was correct. The circuit becomes a 2 input OR gate (notice, it's not a NOR).

Since it is only possible to make a NOR gate, it's output can be the input of an INVERTER to generate an OR function.