#### joker_tony

Joined May 18, 2008
3
i really need someone to me verify my answer. i will be having my exam next week and iam stuck . please iam really stuck on some of this question.
' this is not

'' this is not, not

1. F = B.(A''. B' . C')

2. F = A. (A''+ B' + C')

3. F = A . (B''.C') + A . (B + C')

4. F = (A'' + B'') + A . B

5. F = A'.B'+C'+A.(B'+C')

6. F= A'. B+ D.(C+D') + A. B' . C' . D

7. F = A. C'.D'+(A''.B'.C') + (A''.B')

8. F = A.B'.C' + A'.B.C.D + A'.B.C'+A.B'C.D

1. F=BA'.BC'
2 F=A.AB.AC'
3. F= AB + AC
4. F=A.B

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#### joker_tony

Joined May 18, 2008
3
step by step solution

1. AB'+BB'+BC' (use complements laws B.B'=0)
= AB'+BC'

2. AA.AB'.AC' ( Use idempotency A.A= A)
=A.AB'.AC'

3. AB+AC+AB+AC(use idempotency A+A=A)
=AB+AC

4. (A.B)+A.B ( i think this one is wrong)
=A.AB.AB.B
=A.B

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#### Caveman

Joined Apr 15, 2008
471
1. AB'+BB'+BC' (use complements laws B.B=0)
= AB'+BC'
This is incorrect. B AND B' = 0.

2. AA.AB'.AC' ( Use idempotency A.A= A)
=A.AB'.AC'
This is incorrect as well.

3. AB+AC+AB+AC(use idempotency A+A=A)
=AB+AC
This is incorrect.
4. (A.B)+A.B ( i think this one is wrong)
=A.AB.AB.B=A.B
Incorrect.

I'll explain 1, and you should take it from there:
1. F = B.(A''. B' . C')
First, get rid of the silly '' things A'' = A.
1. F = B.(A . B' . C')
Notice that they are all AND operations, so you can just drop the parenthesis.
1. F = B. A''. B' . C'
You can also swap the order (AND is commutative)
1. F = A''. B. B' . C'
B.B' = 0
1. F = A''. 0 . C'
Anything AND 0 = 0
1. F = 0

#### joker_tony

Joined May 18, 2008
3
y do you have 1. F = B. A''. B' . C'

A'' = A but you have put it back there after you have drop it

you have A''

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#### hgmjr

Joined Jan 28, 2005
9,029
There is some helpful material on Boolean Algebra in the AAC ebook section. There are also some worked examples that may be worth reviewing.

hgmjr

#### sejan

Joined May 24, 2008
7
your no 1 is incorrect,it is 0.
reduction-A(A"+B'+c')
=AA"+AB'+AC'
A+AB'+AC'
no-3,it seems to be ok although my technique is different,i checked with K-map
no-4-
(A"+B")+A.B
=A+B+A.B
=A(1+B)+B
=A+B (1+X=1)
no 5-
A'B'+C'+A(B'+c')
=A'B'+C'+AB'+AC'
=B'(A+A')+C'(1+A)
=B'+C'

I think you can carry on now-if confused use k-map to check

#### Caveman

Joined Apr 15, 2008
471
y do you have 1. F = B. A''. B' . C'

A'' = A but you have put it back there after you have drop it

you have A''
Made a mistake there (cut-and-paste laziness), but the result is the same.