Here we will talk about Digital logic or Boolean logic, So lets decide the syntax: A is some binary input, A’ is negation of A input (NOT of A), + is logical OR operation and, .(dot) is AND operation. Also AB = A.B (AND is implicit here.)
Here we are going to prove/derive Absorption Law:
- A+A.B = A;
- A + A’.B = A+B; // can be read as: (A OR ~AB) ; (A OR (~A AND B)); where ~A is Inverted A;
- A(A+B) = A;
- A(A’+B) = A.B;
Laws are called absorption law as it absorbs one term.
Before we jump to prove Absorption Law, Lets see some other Laws in Boolean Algebra which we may use to prove Absorption Law.
OR Rule: A+1 = 1; A+0 = A; A+A = A;
AND Rules: A.1 = A; A.0 = A; A.A = A;
Double Negation Rule: (A’)’ = A;
Commutative Rule: A+B = B+A; A.B = B.A;
Associative Rule: (A+B)+C = A+(B+C); (A.B)C = A(BC); //don’t confuse BC = B.C;
Distributive Law: A(B+C) = AB+AC; A+BC = (A+B).(A+C) ; // as A(A+B)+C(A+B) = A+AB+AC+BC = A(1+B+C)+BC;
Complements Law: A+A’ = 1; //as 1+0 =1 A.A’ = 0; //as 1.0 = 0;
Reduction Laws: AB+A’B = B; //as B(A+A’) = B.1; (A+B)(A’+B) = B; //as AA’+AB+A’B+BB = 0+B+BB = B;
DeMorgan’s Law: A’+B’ = (AB)’ ;//i.e. ~A+ ~B = ~(AB) A’B’ = (A+B)’ ;//i.e. ~A.~B = ~(A+B)
Lets prove Absorption Law now:
First Law: A+AB = A;
Since A+AB = A(1+B) = A.1 =A;
Second Law: A+A’B = A+B;
We will supply two proves for the second law.
Proof 1: A+A’B = A+B;
Negate both Side, So it becomes (A+A’B)’ = (A+B)’;
Simplify both sides by DeMorgan’s Law, A’.(A’B)’ = A’B’;
Further Simplify left hand side, A’.(A’B)’ = A'((A’)’+B’) = A’.(A+B’) = A’A + A’B’ = 0+A’B’ = A’B’ = Right hand side.
Proof 2 of Second Law: A+A’B = A+B;
Apply Double Negation on Left Hand Side: A+A’B = ((A+A’B)’)’ ;
Expand the inner negation using DeMorgan’s Law,
((A+A’B)’)’ = (A’.(A’B)’ )’ = ( A'((A’)’+B’) )’ = (A'(A+B’))’ =(A’A + A’B’)’ = (0+A’B’)’ = (A’B’)’ = (A’)’ + (B’)’ = A+B;
Third Law: A(A+B) = A;
A(A+B) = AA+AB = A+AB = A(1+B) = A.1 =A;
Fourth Law: A(A’+B) =AB;
A(A’+B) = AA’+AB = 0+AB =AB;
Thanks,
AMAR SINGH
WISESCIENCEWISE 