Question

prove that (X +Y) '' +(X+Y) ''=X+Y in truth table or in any way​

Answer 1

Answer:

like and follow plz

Explanation:

Boolean Algebra Problems

Date: 12/05/97 at 23:30:51

From: John W. Fleming

Subject: Boolean algebra

Could you please help me solve the following boolean algebra problems?

Prove x'y' + x'y + xy' = x' + y'

I have tried this problem several ways and always end up short or with

too many. I cannot get both sides to be equal.

Prove x'y' + x'y + xy' + xy = Identity

On this problem I know what identity means, but I do not understand

what the equation on the right should equal. I know it should be

equal to the identity, but for this problem, what does that mean?

Thank you.

Date: 12/06/97 at 08:18:42

From: Doctor Anthony

Subject: Re: Boolean algebra

(1) Draw up the truth table as follows:

x y x' y' | x'y' + x'y + xy' = f1 | x' + y' = f2 |

----------------------------------------------------------

0 0 1 1 | 1 + 0 + 0 = 1 | 1 + 1 = 1 |

0 1 1 0 | 0 + 1 + 0 = 1 | 1 + 0 = 1 |

1 0 0 1 | 0 + 0 + 1 = 1 | 0 + 1 = 1 |

1 1 0 0 | 0 + 0 + 0 = 0 | 0 + 0 = 0 |

The truth table for the two expressions, f1 and f2, is the same, so

the two represent the same boolean function. i.e. f1 = f2

(2) This is very similar to the last question except that we add xy

to f1 and it is obvious that this makes the total 1 for all possible

combinations of x, x', y, y'.

x y x' y' | x'y' + x'y + xy' + xy = f3

--------------------------------------------

0 0 1 1 | 1 + 0 + 0 + 0 = 1

0 1 1 0 | 0 + 1 + 0 + 0 = 1

1 0 0 1 | 0 + 0 + 1 + 0 = 1

1 1 0 0 | 0 + 0 + 0 + 1 = 1

The truth table shows that f3 = 1 for all x, y, x', y'

What you have called the identity (=1) is here called the universal

set.

Here is an alternatve approach to these problems, not involving truth

tables.

We use fact that xx' = 0, yy' = 0, x+x' = 1 and y+y' = 1

Then starting with the simpler expression we work towards the more

complicated one.

So x' + y' = x'(y+y') + y'(x+x')

= x'y + x'y' + xy' + x'y'

(we only need include one of the x'y')

= x'y + x'y' + xy' (the required expression)

For the second problem we start with:

1 = (x+x')(y+y')

= xy + xy' + x'y + x'y'

In these problems this has proved a quicker method, but truth tables

can be very useful if it is not easy to see in what way we should

'complicate' the easy starting expression.