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# Proof of all Theorems and Postulates of Boolean Algebra

In this article, you will see how to prove all the theorems and postulates available in boolean algebra using the truth table along with algebraic expressions (for some theorem equations).

Let's begin by proving each of the equation's nine theorems and eight postulates one by one.

## Proof of X+X=X

Here is the proof of **X+X=X**

L.H.S. X+X =(X+X).1 [∵ X.1 = X] =(X+X)(X+X') [∵ X+X' = 1] =(X+X.X') [∵ X+YZ = (X+Y)(X+Z)] =X+0 [∵ X.X' = 0] =X [∵ X+0 = X] = R.H.S.

And here is the truth table of the equation **X+X=X**:

X | X+X |
---|---|

0 | 0+0 |

1 | 1+1 |

As X may hold either 0 or 1, we have to check for both possible values. Here in 1^{st} row, **0+0** gives 0, which is equal to X, which
holds 0 in this row.

And in the 2^{nd} row, **1+1** gives 1, which is again equal to X, which holds 1 in this row.

Therefore, **X+X=X** is a correct equation.

## Proof of X.X=X

Here is the proof of **X.X=X**

L.H.S. X.X =(X.X)+0 [∵ X+0 = X] =(X.X)+(X.X') [∵ X.X' = 0] =X(X+X') [∵ X(Y+Z) = (XY+XZ)] =X.1 [∵ X+X' = 1] =X [∵ X.1 = X] = R.H.S.

And this is the truth table for the equation **X.X=X**

X | X.X |
---|---|

0 | 0.0 |

1 | 1.1 |

Here in the 1^{st} row, **0.0** gives 0, which is equal to X in this row.

In the 2^{nd} row, that is **1.1** gives 1, which is again equal to X in this row.

Therefore, the equation **X.X=X** is proved.

## Proof of X+1=1

Below is the proof of the equation **X+1=1**

L.H.S. X+1 =(X+1).1 [∵ X.1=X] =(X+1)(X+X') [∵ X+X'=1] =(X+X'.1) [∵ X+YZ = (X+Y)(X+Z)] =X+X' =1 = R.H.S.

And here is the truth table for the equation **X+1=1**

X | X+1 |
---|---|

0 | 0+1 |

1 | 1+1 |

Here, in the first row, **0+1** gives 1, which is equal to 1. Therefore, the equation X+1=1 is correct in the first row.

And in the second row, **1+1** also gives 1, which is again equal to 1. Therefore, the equation X+1=1 is correct in the second row as well.

Therefore, **X+1=1** is proved.

## Proof of X.0=0

Here is the proof of the equation **X.0=0**

L.H.S. X.0 =(X.0)+0 [∵ X+0 = X] =(X.0)+(X.X') [∵ X.X' = 0] =X(0+X') [∵ X(Y+Z) = XY+XZ] =X.X' =0 = R.H.S.

And below is the truth table for the equation, **X.0=0**

X | X.0 |
---|---|

0 | 0.0 |

1 | 1.0 |

Here, 0.0 gives 0 in the first row, and 1.0 also gives 0 in the second row. Therefore, the equation X.0=0 is proved.

## Proof of (X')'=X

Here is the truth table:

X | X' | (X')' |
---|---|---|

0 | 1 | 0 |

1 | 0 | 1 |

Here, from the above truth table, we can clearly see that (X')' holds the same value as X holds in both the corresponding rows.

Therefore, **(X')' = X**

## Proof of X+(Y+Z)=(X+Y)+Z

Because the equation X+(Y+Z)=(X+Y)+Z has three variables, there are a total of 2^{3} = 8 combinations from 000 to 111. The first digit
represents X, the second represents Y, and the third represents Z. Here is the truth table:

X | Y | Z | X+Y | Y+Z | X+(Y+Z) | (X+Y)+Z |
---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 0 | 1 | 1 | 1 |

0 | 1 | 0 | 1 | 1 | 1 | 1 |

0 | 1 | 1 | 1 | 1 | 1 | 1 |

1 | 0 | 0 | 1 | 0 | 1 | 1 |

1 | 0 | 1 | 1 | 1 | 1 | 1 |

1 | 1 | 0 | 1 | 1 | 1 | 1 |

1 | 1 | 1 | 1 | 1 | 1 | 1 |

As we clearly see, **X+(Y+Z)** holds the same values as **(X+Y)+Z** holds in all the corresponding rows.

As a result, **X+(Y+Z) = (X+Y)+Z** is proved.

## Proof of (X.Y)'=X'+Y'

In the equation **(X.Y)'=X'+Y'**, there are 2 variables; therefore, the total combination will be **2 ^{2} = 4**
from 00 to 11. Here is the truth table of the equation:

X | Y | X' | Y' | X.Y | (X.Y)' | X'+Y' |
---|---|---|---|---|---|---|

0 | 0 | 1 | 1 | 0 | 1 | 1 |

0 | 1 | 1 | 0 | 0 | 1 | 1 |

1 | 0 | 0 | 1 | 0 | 1 | 1 |

1 | 1 | 0 | 0 | 1 | 0 | 0 |

Here in the above truth table, you can see that **(X.Y)'** holds the same value as **X'+Y'** holds in all the corresponding
rows.

As a result, the truth table above identifies (X.Y.)' = X'+Y'.

## Proof of X+XY=X

Below is the proof of the equation **X+XY=X**

L.H.S. X+XY =X.1+XY [∵ X.1 = X] =X(1+Y) [∵ X(Y+Z) = XY+XZ] =X(Y+1) =X.1 =X [∵ X+1 = X] = R.H.S.

The truth table for the equation **X+XY=X** is given here:

X | Y | XY | X+XY |
---|---|---|---|

0 | 0 | 0 | 0 |

0 | 1 | 0 | 0 |

1 | 0 | 0 | 1 |

1 | 1 | 1 | 1 |

Here in this table also, you can see that **X+XY** holds the same value as **X** holds in all the corresponding rows.

Therefore, the equation **X+XY = X** was identified and proved using the above truth table.

## Proof of X(X+Y)=X

The truth table of the equation **X(X+Y)=X** is given below:

X | Y | X+Y | X(X+Y) |
---|---|---|---|

0 | 0 | 0 | 0 |

0 | 1 | 1 | 0 |

1 | 0 | 1 | 1 |

1 | 1 | 1 | 1 |

Here X holds the same values as **X(X+Y)** holds in all the corresponding rows; therefore, the equation **X(X+Y)=X** is proved
using the above truth table.

## Proof of X+0=X

The truth table to show the proof of the equation **X+0=X** is given below.

X | X+0 |
---|---|

0 | 0+0 |

1 | 1+0 |

Here, 0+0 is equal to 0, and 1+0 is equal to 1. Therefore, here we have concluded that **X** holds the same values as **X+0**
holds in all the corresponding rows. Therefore, the equation **X+0=X** is proved using the above truth table.

## Proof of X.1=X

The truth table to show the proof of the equation **X.1=X** is given below:

X | X.1 |
---|---|

0 | 0.1 |

1 | 1.1 |

As such, 0.1 gives 0 as an output or result, and 1.1 gives 1 as an output. Therefore, X holds the same value as X.1 holds in all the corresponding
rows. Therefore, the equation **X.1=X** proved using the above truth table.

## Proof of X+X'=1

The truth table is given below:

X | X' | X+X' |
---|---|---|

0 | 1 | 1 |

1 | 0 | 1 |

In the first row, **X+X' = 0+1** which is equal to 1. And in the second row, **X+X' = 1+0** which is again equal to 1.

Therefore, the equation **X+X'=1** is proved.

## Proof of X.X'=0

The truth table is given here:

X | X' | X.X' |
---|---|---|

0 | 1 | 0 |

1 | 0 | 0 |

In the first row, **X.X' = 0.1** which is equal to 0. And in the second row, **X.X' = 1.0** which is again equal to 0.

Therefore, the equation **X.X'=0** is proved.

## Proof of X+Y=Y+X

Below is the truth table:

X | Y | X+Y | Y+X |
---|---|---|---|

0 | 0 | 0 | 0 |

0 | 1 | 1 | 1 |

1 | 0 | 1 | 1 |

1 | 1 | 1 | 1 |

As we can see from the above truth table, the expression **X+Y** holds the same value as **Y+X** holds in all the corresponding
rows. Therefore, the equation **X+Y=Y+X** proved.

## Proof of XY=YX

The truth table in this case is:

X | Y | XY | YX |
---|---|---|---|

0 | 0 | 0 | 0 |

0 | 1 | 0 | 0 |

1 | 0 | 0 | 0 |

1 | 1 | 1 | 1 |

From the above truth table, we have concluded that **XY** holds the same values as **YX** holds in all corresponding rows.
Therefore, the equation **XY=YX** is proved.

## Proof of X(Y+Z)=XY+XZ

The truth table for this boolean expression is given here. Because the equation X(Y + Z) = XY + XZ contains three variables, X, Y, and Z, we will have a total of eight combinations from 000 to 111, where the first digit represents X, the second represents Y, and the third represents Z.

X | Y | Z | Y+Z | XY | XZ | X(Y+Z) | XY+XZ |
---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 0 | 1 | 0 | 0 | 0 | 0 |

0 | 1 | 1 | 1 | 0 | 0 | 0 | 0 |

1 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

1 | 0 | 1 | 1 | 0 | 1 | 1 | 1 |

1 | 1 | 0 | 1 | 1 | 0 | 1 | 1 |

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

As we can clearly see from the above truth table, **X(X+Z)** holds the same values as the expression **XY+XZ**
in all the corresponding rows. As a result, the postulate equation **X(Y+Z)=XY+XZ** is proved.

## Proof of X+YZ=(X+Y)(X+Z)

Here also, we have a total of 3 variables, which means we will have a total of 8 combinations as shown in the following truth table:

X | Y | Z | YZ | X+Y | X+Z | X+YZ | (X+Y)(X+Z) |
---|---|---|---|---|---|---|---|

0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |

0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 |

0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 |

0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 |

1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 |

1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 |

1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 |

Here also, the expression **X+(YZ)** holds the same values as **(X+Y)(X+Z)** holds in all the corresponding rows. As a result,
**X+YZ=(X+Y)(X+Z)** is proved.

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