Let S = {O, 1}, and define operations + and · on S by the following tables:

a. Show that the elements of S satisfy the following properties:
(i) the commutative law for +
(ii) the commutative law for ·
(iii) the associative law for +
(iv) the associative law for ·
(v) the distributive law for + over ·
(vi) the distributive law for · over +
b. Show that 0 is an identity element for + and that 1 is an identity element for ·.
c. Define 0 ¯ = 1 and 1 ¯ = 0 . Show that for every a in S. a + a ¯ = 1 and a ⋅ a ¯ = 0 . It follows from parts (a)—(c) that S is a Boolean algebra with the operations + and ·.

(a)

To prove:

Show that the elements of S satisfies:

1. The commutative law for +
2. The commutative law for ⋅
3. The associative law for +
4. The associative law for ⋅
5. The distributive law for + over ⋅
6. The distributive law for ⋅ over +

Given information:

Let S={0,1}, and define operations + and ⋅ on S by the following tables:

Proof:

(i) To proof: x+y=y+x

PROOF:

x+y=1 if at least one of x and y is 1.

x+y=0 if x and y are both 0.

First case: x = 0 and y = 0

x+y=0+0=0=0+0=y+x

Second case: x = 0 and y = 1

x+y=0+1=1=1+0=y+x

Third case: x = 1 and y = 0

x+y=1+0=1=0+1=y+x

Fourth case: x = 1 and y = 1

x+y=1+1=1=1+1=y+x

Thus, we note that the commutative law x+y=y+x is true in each case.

(ii) To proof: x⋅y=y⋅x

PROOF:

x⋅y=1 if x and y are both 1.

x⋅y=0 if at least one of x and y is 0.

First case: x = 0 and y =0

x⋅y=0⋅0=0=0⋅0=y⋅x

Second case: x = 0 and y = 1

x⋅y=0⋅1=0=1⋅0=y⋅x

Third case: x = 1 and y = 0

x⋅y=1⋅0=0=0⋅1=y⋅x

Fourth case: x = 1 and y = 1

x⋅y=1⋅1=1=1⋅1=y⋅x

Thus we note that the commutative law x⋅y=y⋅x is true in each case.

(iii) To proof: (x+y)+z=x+(y+z)

PROOF:

x+y=1 if at least one of x and y is 1.

x+y=0 if x and y are both 0.

First case: x = 0 and y = 0 and z = 0

(x+y)+z=(0+0)+0=0+0=0

x+(y+z)=0+(0+0)=0+0=0

Second case: x = 0 and y = 0 and z = 1

(x+y)+z=(0+0)+1=0+1=1

x+(y+z)=0+(0+1)=0+1=1

Third case: x = 0 and y = 1 and z = 0

(x+y)+z=(0+1)+0=1+0=1

x+(y+z)=0+(1+0)=0+1=1

Fourth case: x = 0 and y = 1 and z = 1

(x+y)+z=(0+1)+1=1+1=1

x+(y+z)=0+(1+1)=0+1=1

Fifth case: x = 1 and y = 0 and z = 0

(x+y)+z=(1+0)+0=1+0=1

x+(y+z)=1+(0+0)=1+0=1

Sixth case: x = 1 and y = 0 and z = 1

(x+y)+z=(1+0)+1=1+1=1

x+(y+z)=1+(0+1)=1+1=1

Seventh case: x = 1 and y = 1 and z = 0

(x+y)+z=(1+1)+0=1+0=1

x+(y+z)=1+(1+0)=1+1=1

Eight case: x = 1 and y = 1 and z = 1

(x+y)+z=(1+1)+1=1+1=1

x+(y+z)=1+(1+1)=1+1=1

Thus we note that the associative law (x+y)+z=x+(y+z) is true in each case.

(iv) To proof: (x⋅y)⋅z=x⋅(y⋅z)

PROOF:

x⋅y=1 if x and y are both 1.

x⋅y=0 if at least one of x and y is 0.

First case: x = 0 and y = 0 and z = 0

(x⋅y)⋅z=(0⋅0)⋅0=0⋅0=0

x⋅(y⋅z)=0⋅(0⋅0)=0⋅0=0

Second case: x = 0 and y = 1 and z = 0

(x⋅y)⋅z=(0⋅0)⋅1=0⋅1=0

x⋅(y⋅z)=0⋅(0⋅1)=0⋅0=0

Third case: x = 0 and y = 1 and z = 0

(x⋅y)⋅z=(0⋅1)⋅0=0⋅0=0

x⋅(y⋅z)=0⋅(1⋅0)=0⋅0=0

Fourth case: x = 0 and y = 1 and z = 1

(x⋅y)⋅z=(0⋅1)⋅1=0⋅1=0

x⋅(y⋅z)=0⋅(1⋅1)=0⋅1=0

Fifth case: x = 1 and y = 0 and z = 0

(x⋅y)⋅z=(1⋅0)⋅0=0⋅0=0

x⋅(y⋅z)=1⋅(0⋅0)=1⋅0=0

Sixth case: x = 1 and y = 0 and z = 1

(x⋅y)⋅z=(1⋅0)⋅1=0⋅1=0

x⋅(y⋅z)=1⋅(0⋅1)=1⋅0=0

Seventh case: x = 1 and y = 1 and z = 0

(x⋅y)⋅z=(1⋅1)⋅0=1⋅0=0

x⋅(y⋅z)=1⋅(1⋅0)=1⋅0=0

Eight case: x = 1 and y = 1 and z = 1

(x⋅y)⋅z=(1⋅1)⋅1=1⋅1=1

x⋅(y⋅z)=1⋅(1⋅1)=1⋅1=1

Thus we note that the associative law (x⋅y)⋅z=x⋅(y⋅z) is true in each case.

(v) To proof: x+(y⋅z)=(x+y)⋅(x+z)

PROOF:

x+y=1 if at least one of x and y is 1

(b)

To prove:

0 is an identity element for + and that 1 is an identity element for ⋅.

(c)

To prove:

For every a in S, a+a¯=1 and a⋅a¯=0.