Boolean Algebra Proofs Postulates and Theorems (Part 2)

Boolean Algebra Postulates and Theorems (Part 2):

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Commutative Law:

x + y = y + x;
x .  y  = y . x;


Associative Law:

x + (y + z) = (x + y) + z;
x .  (y .  z) = (x  . y) .  z;


Distributive law:

x . (y + z) = x  .  y  + x  .  z;
x + (y . z) = (x + y) . (x + z);


De Morgan’s Law:

(x + y)’ = x’ .  y’;
(x .  y)’ = x’ + y’;


Absorption Law:

x + x  .  y = x;
x . (x + y) = x;


Absorption Law Proof:

x + x.y = x
L.H.S =>
= x.(1 + y)   [Distributive Law]
= x.1         [We know, x + 1 = x]
= x           [We know, x . 1 = x]
= R.H.S

Boolean Algebra Proofs Postulates and Theorems (Part 1)

Boolean Algebra Postulates and Theorems (Part 1):


First familiarize with truth tables so it’ll be easier to understand.

 

x + 0 = x

here only two possible states of x, 0 remains constant

x = 0
x = 1

So,

false OR false is always false
0 + 0 = 0
true OR false is always true
1 + 0 = 1

So, from this we can see whatever the value of x is, the output is always equal to x.


 

 x . 1 = x

here only two possible states of x, 1 remains constant

x = 0
x = 1

So,

false AND true is always false
0 . 1 = 0
true AND true is always true
1 . 1 = 1

So, from this we can see whatever the value of x is, the output is always equal to x.


 

 x + 1 = 1

here only two possible states of x, 1 remains constant

x = 0
x = 1

So,

false OR true is always true
0 + 1 = 1
true OR true is always true
1 + 1 = 1

So, from this we can see no matter the value of x, OR with 1 (true) always gives a 1 (true) value.


 

 x . 0 = 0

here only two possible states of x, 0 remains constant

x = 0
x = 1

So,

false AND false is always false
0 . 0 = 0
true AND false is always false
1 . 0 = 0

So, from this we can see no matter the value of x, AND with 0 (false) always gives a 0 (false) value.


 

 x + x’ = 1

x = 0, x' = 1
x = 1, x' = 0

So,

false OR true is always true
0 + 1 = 1
true OR false is always true
1 + 0 = 1

 

 x . x’ = 0

x = 0, x' = 1
x = 1, x' = 0

So,

false AND true is always false
0 . 1 = 0
true AND false is always false
1 . 0 = 0

 

 x + x = x

x = 0, x = 0
x = 1, x = 1

So,

false OR false is always false
0 . 0 = 0
true OR true is always true
1 . 1 = 1

So, we can whatever the value of x is, that is our output.


 

x . x = x

x = 0, x = 0
x = 1, x = 1

So,

false AND false is always false
0 . 0 = 0
true AND true is always true
1 . 1 = 1

So, again we can whatever the value of x is, that is our output.


 

 (x’)’ = x

x = 0, x' = 1, (x')' = 0
x = 1, x' = 0, (x')' = 1

So, we can see complementing twice gives the original value.