Discrete Mathematics..

460 Boolean AlgebraNow imagine that we have a circuit (assumed to include a suitable power source)which contains a switch A. We denote the state of the switch by the variable xwhere x = 0 if A is open and x = 1 if A is closed.Consider now a circuit which contains two switches A 1 and A 2 connected asshowninthediagrambelow.Switches connected to each other in this way are said to be in series. It is clear thatcurrent will flow across this section of a circuit only if both switches A 1 and A 2are closed. Let x 1 and x 2 be variables denoting the states of switches A 1 and A 2respectively. (In each case 0 denotes open and 1 denotes closed.) Let f (x 1 , x 2 )be a function which has the value 1 for values of x 1 and x 2 which allow currentto flow and 0 otherwise. Thus f :{0, 1} 2 →{0, 1} and the value of f (x 1 , x 2 ) forall possible values of x 1 and x 2 is given in the table below.x 1 x 2 f (x 1 , x 2 )0 0 00 1 01 0 01 1 1We can now see that f is the familiar function f (x 1 , x 2 ) = x 1 x 2 wherex 1 and x 2 are variables whose domain is the two-element Boolean algebra({0, 1}, ⊕, ∗,¯, 0, 1).Two switches may alternatively be connected in parallel, the arrangement shownin the diagram below.For current to flow around a circuit containing a power source and just twoswitches connected in this way, it is necessary that one or both of the switchesare closed. Defining x 1 and x 2 as before and g(x 1 , x 2 ) exactly as we defined