Page 74 - Computer Science Class 11 Without Functions
P. 74

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)
                       = 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-XI
   69   70   71   72   73   74   75   76   77   78   79