Question

In addition to the basic laws we discussed in this section, there are two important theorems, called DeMorgan’s theorems:

$\overline{\mathbf{A}\mathbf{+}\mathbf{B}}\mathbf{=}\overline{\mathbf{A}}\mathbf{·}\overline{\mathbf{B}}$and$\overline{\mathbf{A}\mathbf{·}\mathbf{B}}\mathbf{=}\overline{\mathbf{A}}\mathbf{+}\overline{\mathbf{B}}$

Prove DeMorgan’s theorems with a truth table of the form

A	B	$\overline{\mathbf{A}}$	$\overline{\mathbf{B}}$	$\overline{\mathbf{A}\mathbf{+}\mathbf{B}}$	$\overline{\mathbf{A}}\mathbf{·}\overline{\mathbf{B}}$	$\overline{\mathbf{A}\mathbf{·}\mathbf{B}}$	$\overline{\mathbf{A}}\mathbf{+}\overline{\mathbf{B}}$
0	0	1	1	1	1	1	1
0	1	1	0	0	0	1	1
1	0	0	1	0	0	1	1
1	1	0	0	0	0	0	0

Short answer

Prove is provided below

Step-by-step solution

Define DeMorgan’s 1st law

DeMorgan’s First law states that the complement of the sum of all the terms is equal to the product of the complements of each term.

$\overline{A+B}=\overline{A}·\overline{B}$

Define DeMorgan’s  2nd law

DeMorgan’s Second law state that the complement of the product of all terms is equal to the sum of complements of each term

$\overline{A·B}=\overline{A}+\overline{B}$

Prove 1st theorem A+B=A·Busing the truth table

$\overline{A+B}=\overline{A}·\overline{B}$

Proof:

Using complementarity laws, $A+\overline{A}=1andA·\overline{A}=0$

role="math" localid="1655189870848" $$\begin{gathered} \left(A+B\right)+\left(\overline{A+B}\right)=1\&\left(A+B\right)·\left(\overline{A+B}\right)=0......\left(1\right) \\ \left(A+B\right)+\left(\overline{A}·\overline{B}\right)=0\&\left(A+B\right)·\left(\overline{A}·\overline{B}\right)=0......\left(2\right) \end{gathered}$$

First, we have to prove $\left(A+B\right)+\left(\overline{A}·\overline{B}\right)=0$

$$\begin{gathered} =\left(\left(A+B\right)+\overline{A}\right)·\left(\left(A+B\right)+\overline{B}\right)\left(A+BC=\left(A+B\right)·\left(A+C\right)\right) \\ =\left(A+\overline{A}+B\right)·\left(A+B+\overline{B}\right)\left(CommutativeLaw\right) \end{gathered}$$

$$\begin{gathered} =\left(1+B\right)·\left(A+1\right)\left(A+\overline{A}=1\right) \\ =1·1\left(1+A=1\right) \\ =1 \end{gathered}$$

Now, we prove $\left(A+B\right)·\left(\overline{A}·\overline{B}\right)=0$

$$\begin{gathered} \left(A+B\right)·\left(\overline{A}+\overline{B}\right)=0 \\ =\overline{A}+\overline{B}·\left(A+B\right)(CommutativeLaw) \\ =\overline{A}\overline{B}A+\overline{A}\overline{B}B(DistributiveLaw) \\ =A\overline{A}\overline{B}+\overline{A}B\overline{B}(CommutativeLaw) \\ =0·\overline{B}+\overline{A}·0\left(A·\overline{A}=0\right) \\ =0+0 \\ =0 \end{gathered}$$

Hence proved LHS=RHS

Prove 1st theorem  A·B=A+Busing the truth table

$\overline{A·B}=\overline{A}+\overline{B}$

Proof:

Using complementarity laws, $A+\overline{A}=1,A·\overline{A}=0$

$$\begin{gathered} AB+\overline{A}\overline{B}=1\&\left(AB\right)·\left(\overline{A}\overline{B}\right)=0......\left(1\right) \\ AB+\left(\overline{A}+\overline{B}\right)=1\&\left(AB\right)·\left(\overline{A}+\overline{B}\right)=0......\left(2\right) \end{gathered}$$

First, we prove $AB+\left(\overline{A}+\overline{B}\right)=1$

$$\begin{gathered} =AB+\left(\overline{A}+\overline{B}\right) \\ =\left(\overline{A}+\overline{B}\right)+\left(AB\right)(Commutativelaw) \\ =\left(\overline{A}+\overline{B}+A\right)·\left(\overline{A}+\overline{B}+B\right)\left(A+BC=\left(A+B\right)·\left(A+C\right)\right) \\ =\left(1+\overline{B}\right)·\left(1+\overline{A}\right) \\ =1·1 \\ =1 \end{gathered}$$

LHS=RHS

Now we prove $$\begin{gathered} AB·\left(\overline{A}+\overline{B}\right)=0 \\ =AB\overline{A}+AB\overline{B}(Distributivelaw) \\ =0·B+A·0 \\ =0+0 \\ =0 \end{gathered}$$

Hence proved LHS = RHS