Page 86 - Computer Science V2.0 Class 11

Let us verify the following Absorption Law, using truth Table 3.6.
a + a ● b = a

a b a ● b a + a ● b
0 0 0 0
0 1 0 0
1 0 0 1
1 1 1 1

Table 3.6: a + a.b = a
As column 1 (variable a) and column 4 (a + a ● b) are same, we have verified the first of the two absorption laws.
Similarly, you may verify the other absorption law and the following theorems.
Theorem 4: Idempotent Law: ∀ a ∈ S
(i) a + a = a

(ii) a ● a = a
Theorem 5: Involution: ∀ a ∈ S, (a')' = a
Theorem 6: Associative Law: ∀, a, b, c ∈ S,
(i) (a + b) + c = a + (b + c)
(ii) (a ● b) ● c = a ● (b ● c)

Theorem 7: De Morgan's laws: ∀ a, b ∈ S
(i) (a + b)' = a' ● b'
(ii) (a ● b)' = a' + b'
Proof: (i) To prove (a + b)' = a' ● b', we shall show that a' ● b' is the complement of (a + b)

Find we prove: (a + b) + (a' ● b') = 1
(a + b) + (a' ● b')
= a + (b + (a' ● b')) (Using associativity property)
= a + ((b + a') ● (b + b')) (Using distributive property of +)
= a + (b + a') ● 1 (Using complementarity property)

= a + (b + a') (Using multiplicative identity)
= a + (a'+ b) (Using commutative property)
= (a + a') + b (Using associativity property)
= 1 + b (Using complementarity property)

= 1 (Using Theorem 2 (i))
Next, we prove: (a + b) ● (a'● b') = 0
(a + b) ● (a' ● b')
= a ● (a' ● b') + b ● (a' ● b') (Using distributive property of ●)
= (a ● a') ● b' + b ● (a' ● b') (Using associative property)

= 0 ● b' + b ● (a' ● b') (Using complementarity property)
= 0 + b ● (a' ● b') (Using theorem 2 (ii))
= b ● (a' ● b') (Using additive identity)

= b ● (b' ● a') (Using commutative property)

72 Touchpad Computer Science (Ver. 2.0)-XI