prove that in the boolenan algebra B, the additive identive identify i.e. '0' of B is unique.
Answers
A non-empty set B = (a, b, c …..) with
Note: some authors include associative law in the definition of Boolean Algebra.
Theorem VII. The complement of a \in B is uique.
Proof: let there are two complements a’ and a’’.
Then a + a’ = 1, a.a’ = 0
a + a’’ = 1, a.a’’ = 0
now a’ = 1.a’ = (a + a’’).a’
= a.a’ + a’’.a’
= 0 + a’’ . a’ = a.a’’ + a’’ .a’
=a’’.(a + a’) = a’’.1
= a’’
Here a’ is unique.
Theorem VIII (a’)’ = a
Proof : (a’)’ = a.a’ + (a’)’
= {a + (a’)’}. {a’ + (a’)’} …(B II)
= {a + (a’)’}.1
= {a + (a’)’}.{a + a’}
= a + (a’).a’ = a+ 0
=a
The theorem is direct result of B IV
Theorem IX: complements of identities
0’ = 1 and 1’ =0.
Proof: the proof follows from B III and B IV as
1 + 0 = 1 and 1.0 =0
Theorem X: de morgan’s law.
(a +b)’ = a’.b’
And (a.b)’ = a’ + b’
Proof:
(a’.b’).(a + b) = (a’.b’) . a + (a’.b’).b
= a’.(b’.a) + a’.(b’.b) (Th. VI)
= a’.(a.b’) + a’.0
= (a’.a).b’ + 0 = 0.b’ + 0
= 0
Hence a’.b’ is the complement of (a + b)
Also (a.b).(a’ + b’) = {(a.b).a’} + {(a.b).b’}
= {a’. (a.b)} + a. {(b.b’)}
= {(a’.a).b} {a.0}
= 0.b + a.0 = 0 + 0 = 0
And (a.b) + (a’ + b’) = (a’ + b’) + (a.b)
= {(a’ + b’) + a}.{(a’ + b’) + b}
= {a + (a’ + b’)}.{a’ + (a’ + (b’ + b)}
= {(a + a’) + b’}. {a’ + 1}
= (1 + b’). (a’ + 1) = 1.1 = 1
Hence a’ +b’ is the complement of a.b.
Duality
Any result that is obtained from the axioms of Boolean Algebra remains true if + and \bullet are interchanged along with the interchange of 0 and 1 throughout the statement of the results.