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Boolean Algebra
This property of Boolean algebra state that all binary expressions remain valid when following two steps are performed:
Step 1: Interchange OR and AND operators.
Step 2: Replace all 1’s by 0’s and 0’s by 1’s.
Adding a number to ‘0’ to itself
P1. Postulate:-
From duality of P2
P2. Postulate:-
Sum of a number and its complement from a Set is ‘1’
P3. Postulate:-
From duality of P3
P4. Postulate:-
From commutative property of binary numbers we have
P5. Postulate:-
From duality of P5
P6. Postulate:-
From distributive property of binary numbers we have
P7. Postulate:-
From duality of P7
P8. Postulate:-
T1. Theorem:-
x + x = (x + x)*1= (x + x)(x + x’)
From P8, x + xx’ = x
From duality of T1
T2. Theorem:-
T3. Theorem:-
x + 1= (x + 1).1= (x +1)*(x + x’)
From P8, (x + 1*x’)= (x + x’)= 1
From duality of T3
T4. Theorem:-
T5. x + (y + z) = (x + y) + z
From duality of T5
T6. Theorem:-
T7. Theorem:-
From duality of T10
T11. Theorem:-
T8. Theorem:-
T9. Theorem:-
T10. Theorem:-