Using Axioms & other Theorems to prove something 2. Theorems can also be proved

Question

Using Axioms & other Theorems to prove something 2. Theorems can also be proved through a series of Boolean transformations. For example, Theorem 3a can be solved the following way (trying to show yx+xx): = x(y + 1) = x(1) II distributivity axiom I Theorem 2a Il identity axiom yx + x Now, try to prove Theorem 3b using axioms and other theorems (assume all other theorems have been proven already). Be sure to make a comment next to each step telling us what theorem/axiom you are using. "It may be helpful to watch the Boolean Functions video ahead of time. (y +x)x

Explanation / Answer

Ans 1

Therom 1(a)

X+X = X

we take l.h.s

(X+X) .1 // BY 2b

(X+X) (X+X')

(X+XX')

(X+0)

= X R.H.S

Therom 1(b)

XX = X

take L.H.S

X.X + 0

X.X + X.X'

X (X+X')

X .1 // by 2a

X R.H.S // indentity axiom

Ans 2

Therom 2

X+1 = 1

= 1. (X+1)

(X+X') (X+1)

X + X' . 1

X + X'

= 1 R.H.S

Ans 3

Therom 3

yx+x = x // therom 3a

We take L.H.S

yx + x .1 // by 2b

x (y + 1)

x . 1 // by 2(a)

x R.H.S

Ans 4

Therom 4

(x')' = x

in this we take a boolean table

if x= 0, 0 , 1 , 1

then

x x' (x')'

0 1 0

0 1 0

1 0 1

1 0 1

then we show that (X')' = X

Ans 5

Therom 5

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

if we enter in this , X= 1 , Y and Z = 0 and then prove

(1+0)+0 = 1+(0+0)

1 = 1

we show that it is prove

Ans 6

Therom 6

(x+y)' = x'y'

it is sufficient (x+y) + x'.y' = 1

L.H.S

y+(x+ x'.y')

y+x+y'

(y+y') + x

1+x

= 1 R.H.S

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