Question 1190816: Simplify the following Boolean Functions
1. F = AB(A’+BC)’
2. F=AB+AB’(A’C’)’
3. F=ABC + A’ + AB’C
4. F = A’B’C’ + A’B’C + A’C’
5. F = A + A’B
6. F=(A + B)(A+C)
Can you please explain this homework question step-by-step? Thank you very much!
Answer by math_tutor2020(3817)   (Show Source):
You can put this solution on YOUR website!
Before we get started, here is a list of Boolean Algebra rules that you should memorize or have on a reference chart nearby.
| Name | AND form | OR form | | Identity Law | 1*A = A | A+0 = A | | Null Law | 0*A = 0 | A+1 = 1 | | Idempotent Law | A*A = A | A+A = A | | Inverse Law | A*A' = 0 | A+A' = 1 | | Commutative Law | A*B = B*A | A+B = B+A | | Associative Law | A*(B*C) = (A*B)*C | A+(B+C) = (A+B)+C | | Distributive Law | A+B*C = (A+B)*(A+C) | A*(B+C) = A*B+A*C | | Absorption Law | A*(A+B) = A | A+A*B = A | | De Morgan's Law | (A*B)' = A' + B' | (A+B)' = A'*B' |
The star or asterisk symbol is sometimes omitted.
This means something like A*B is the same as AB
In some Boolean algebra textbooks, the tickmark is replaced with a horizontal bar overhead.
Eg: 
I'll use the tickmark notation here.
In Boolean algebra, there are only two "numbers" and they are 0 and 1. I put "numbers" in quotes because they aren't treated like numbers in the traditional sense.
They are more like symbolic placeholders for true and false.
0 = false
1 = true
I'm assuming you have taken logic class as a prerequisite. It would help reinforce the ideas I'll mention.
The multiplication sign means "and", so we have a conjunction here.
A*B = A and B
A*B = 1 if and only if A = B = 1. Otherwise, A*B = 0
The plus sign means "or". We have a disjunction.
A+B = 0 if and only if A = B = 0. Otherwise, A+B = 1.
So because of the disjunction rule mentioned, this means we have the really strange rule that A+B = 1+1 = 1.
Once again, we're not treating the '1's like the traditional sense, and the plus sign isn't used in the way you'd expect.
As long as you keep in mind we're not really working with numbers, but rather true/false values, then you should be fine.
The rules mentioned above behave almost exactly as you'd expect in a regular algebra class.
Something like A*(B+C) = A*B+A*C is the common distributive rule you learn in high school.
Then there's this weird "AND" form where it says A+BC = (A+B)(A+C) which basically has us flip each * to + and vice versa.
I.e.
A*(B+C) = (A*B) + (A*C)
flips to
A+(B*C) = (A+B) * (A+C)
which hopefully shows the connection between the "AND"/"OR" forms.
One last rule to keep in mind, that I couldn't fit with the table above, is the Involution Law which says (A')' = A
This is because the tickmarks effectively flip the truth value. Eg: if A = 0 then A' = 1. Applying another tickmark will flip things again.
Hopefully these concepts aren't entirely brand new, but rather a very quick crash course refresher.
Now onto the math problems.
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Problem 1
F = AB(A'+BC)'
F = AB( (A')'*(BC)' ) .... De Morgan's Law
F = AB( A*(BC)' ) .... Involution Law
F = ABA*(BC)' .... Associative Law
F = AAB*(BC)' .... Commutative Law
F = AB*(BC)' .... Idempotent Law
F = AB*(B' + C') .... De Morgan's Law
F = ABB' + ABC' .... Distributive Law
F = A*0 + ABC' .... Inverse Law
F = 0 + ABC' .... Null Law
F = ABC' .... Identity Law
If you want, you can make a truth table or Boolean table to help confirm the answer.
| A | B | C | C’ | AB | A' | BC | A’+BC | (A'+BC)' | AB(A'+BC)' | ABC’ | | 1 | 1 | 1 | 0 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | | 1 | 1 | 0 | 1 | 1 | 0 | 0 | 0 | 1 | 1 | 1 | | 1 | 0 | 1 | 0 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | | 1 | 0 | 0 | 1 | 0 | 0 | 0 | 0 | 1 | 0 | 0 | | 0 | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 0 | 0 | 0 | | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 1 | 0 | 0 | 0 | | 0 | 0 | 0 | 1 | 0 | 1 | 0 | 1 | 0 | 0 | 0 |
The last two columns match up perfectly.
This proves that AB(A'+BC)' = ABC' since the inputs A,B,C lead those two outputs to produce the same 1's and 0's.
Both sides are identical or interchangeable.
Note that the first column for A is four 1's followed by four 0's.
Column B consists of two identical blocks of "1100"
Column C consists of four identical strings of "10"
This configuration helps set up all possible combos of '1's and '0's.
========================================
Problem 2
F = AB + AB'(A'C')'
F = AB + AB'((A')'+(C')') ... De Morgan's Law
F = AB + AB'(A+C) ... Involution Law
F = (AB + AB')*(AB+(A+C)) ... Distributive Law (use the "AND" form)
F = (A(B + B'))*(AB+(A+C)) ... Distributive Law
F = (A*(1))*(AB+(A+C)) ... Inverse Law
F = A(AB+(A+C)) ... Identity Law
F = A((AB+A)+C) ... Associative Law
F = A((A+AB)+C) ... Commutative Law
F = A(A+C) ... Absorption Law
F = A*A+A*C ... Distributive Law
F = A+AC ... Idempotent Law
F = A ... Absorption Law
Once again you can make a truth table or Boolean table to confirm the answer. I'll let you do this step.
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Problem 3
F = ABC + A' + AB'C
F = A' + ABC + AB'C ... Commutative law
F = A' + ACB + ACB' ... Commutative law
F = A' + AC(B+B') ... Distributive law
F = A' + AC(1) ... Inverse law
F = A' + AC ... Identity law
F = (A' + A)*(A' + C) ... use the "AND" form of the distributive law
F = (1)*(A' + C) .... Inverse Law
F = A' + C ... Identity Law
Like with problem 2, I'll let you create the verification truth table.
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Problem 4
F = A'B'C' + A'B'C' + A'C'
F = A'B'C' + A'C' ..... Idempotent Law
F = A'(B'C' + C') ..... Distributive Law
F = A'(C'B' + C') ..... Commutative Law
F = A'(C'(B' + 1)) ..... Distributive Law
F = A'(C'(1)) ..... Null Law
F = A'C' ..... Identity Law
Or a slightly different way to do this is:
F = A'B'C' + A'B'C' + A'C'
F = A'B'C' + A'C' ..... Idempotent Law
F = A'(B'C' + C') ..... Distributive Law
F = A'(C' + B'C') ..... Commutative Law
F = A'(C' + C'B') ..... Commutative Law
F = A'C' ..... Absorption Law
========================================
Problem 5
F = A + A'B
F = (A + A')*(A + B) ... use the "AND" form of the distributive law
F = (1)*(A+B) ... Inverse Law
F = A+B .... identity law
Boolean table confirmation
| A | B | A’ | A’B | A+A’B | A+B | | 1 | 1 | 0 | 0 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 | 1 | | 0 | 1 | 1 | 1 | 1 | 1 | | 0 | 0 | 1 | 0 | 0 | 0 |
This proves that A+A'B = A+B
========================================
Problem 6
F = (A+B)(A+C)
F = (A+B)*A+(A+B)*C ... Distributive Law
F = A*(A+B)+C*(A+B) ... Commutative Law
F = A*A+A*B+C*A+C*B) ... Distributive Law
F = A+AB + CA+CB ... Idempotent Law
F = A+AB + AC+BC ... Commutative Law
F = A + AC+BC ... Absorption Law
F = A+BC ... Absorption Law
Or you can use the one-step shortcut of this
F = (A+B)(A+C)
F = A+BC
when you use the "AND" form of the distributive law
Boolean table confirmation
| A | B | C | BC | A+B | A+C | (A+B)(A+C) | A+BC | | 1 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | | 1 | 1 | 0 | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 1 | 0 | 1 | 1 | 1 | 1 | | 1 | 0 | 0 | 0 | 1 | 1 | 1 | 1 | | 0 | 1 | 1 | 1 | 1 | 1 | 1 | 1 | | 0 | 1 | 0 | 0 | 1 | 0 | 0 | 0 | | 0 | 0 | 1 | 0 | 0 | 1 | 0 | 0 | | 0 | 0 | 0 | 0 | 0 | 0 | 0 | 0 |
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