DeMorgan’s Theorem

DeMorgan’s Theorem is mainly used to solve the various Boolean algebra expressions. There are two DeMorgan’s Theorems. They are described below in detail.

DeMorgan’s First Theorem

According to DeMorgan’s first theorem a NOR gate is equivalent to a bubbled AND gate.

The Boolean expressions for the bubbled AND gate can be expressed by the equation shown below.

For NOR gate, the equation is

demorgan's-theorem-eq1

For the bubbled AND gate the equation is

demorgan's-theorem-eq2

As the NOR and bubbled gates are interchangeable, i.e., both gates have exactly identical outputs for the same set of inputs.

Therefore, the equation can be written as shown below.

demorgan's-theorem-eq3

This equation (1) or identity shown above is known as DeMorgan’s Theorem. The symbolic representation of the theorem is shown in the figure below.

De-morgan's-theorem-fig-1DeMorgan’s Second Theorem

DeMorgan’s Second Theorem states that the NAND gate is equivalent to a bubbled OR gate.

The Boolean expressions for the NAND gate is given by the equation shown below.

demorgan's-theorem-eq4

The Boolean expressions for the bubbled OR gate is given by the equation shown below.

demorgan's-theorem-eq5

Since, NAND and bubbled OR gates are interchangeable, i.e., both gates have identical outputs for the same set of inputs. Therefore, the equations become as given below.

demorgan's-theorem-eq6

This identity or equation (2) shown above is known as DeMorgan’s Second Theorem.

The symbolic representation of the theorem is shown in the figure below.

DeMorgan's-theorem-fig-2The Bubbled OR Gate

The logic circuit of the bubbled OR gate is shown below.

De-morgan's-theorem-fig-3The truth table for bubbled OR gate is shown below.

ABZ
001
011
101
110

In this, both the inputs are inverted before they are applied to an OR gate. The output of the a bubbled OR gate can be derived from its logic circuit and can be expressed by the equation shown below.

demorgan's-theorem-eq7

Here are the results when the logic circuit of bubbled OR gate when all the possible sets of inputs are applied such as 00, 01, 10 or 11.

For AB: 00

demorgan's-theorem-eq8

For AB: 01

demorgan's-theorem-eq9

For AB: 10

demorgan's-theorem-eq10

For AB: 11

demorgan's-theorem-eq11

The truth table for the bubbled AND gate is exactly identical to the truth table of a NAND gate. Hence, NAND and bubbled OR gate is interchangeable.

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