If the output `y^(')` of NOR gate is used as the input of NOT gate (made from NOR gate by joining two inputs), as shown in Fig., then we get back an OR gate.
The output of gate I is, `y^(')=bar(A+B)=barA.barB`
(From De Morgan's theorem)
The output of IIis,
`y=bar(y^('))=bar(barA.barB)=bar(barA)+bar(barB)=A+B`
which is the Boolean expression for OR gate. The truth table of the combination will be as shown in Fig , which is the truth table of OR gate.
