Question

Simplify:
a'c+a'b+ab'c'+bc

Answer 1

Answer:

I have added simplification by Bollean algebra

and sorry for long answer







How can I simplify the following Boolean expression? A'C+A'.B+AB'C+BC

Algebra

.

Explanation:

Converting the primes to bars:

f=¯¯¯AC+¯¯¯AB+A¯¯¯BC+BC

Let's start with an empty 3 variable Karnaugh map:

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0000A0000∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

Enter the 1s corresponding to the first term into the map:

f=¯¯¯AC+¯¯¯AB+A¯¯¯BC+BC

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0110A0000∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

Enter the 1s corresponding to the second term into the map:

f=¯¯¯AC+¯¯¯AB+A¯¯¯BC+BC

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0111A0000∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

Enter the 1s corresponding to the third term into the map:

f=¯¯¯AC+¯¯¯AB+A¯¯¯BC+BC

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0111A0100∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

Enter the 1s corresponding to the fourth term into the map:

f=¯¯¯AC+¯¯¯AB+A¯¯¯BC+BC

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0111A0110∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

We are ready to begin minimization with the following map:

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0111A0110∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

Please observe that the 4 1s in the center tell us that the function they are only dependent on C:

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0111A0110∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

fmimimized=C

Add the term ¯¯¯AB for the pair of 1s in the upper right:

¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯∣∣ ∣ ∣∣ ¯¯¯B¯¯¯C¯¯¯BCBCB¯¯¯C¯¯¯A0111A0110∣∣ ∣ ∣∣−−−−−−−−−−−−−−−−−−−−−−−

fmimimized=C+¯¯¯AB

Starting with the function with notation converted from primes to bars:

f=¯¯¯AC+¯¯¯AB+A¯¯¯BC+BC

Group all of the terms containing C together:

f=¯¯¯AB+¯¯¯AC+A¯¯¯BC+BC

Remove a factor of C:

f=¯¯¯AB+C(¯¯¯A+A¯¯¯B+B)

AND the ¯¯¯A term with 1 in the form of (B+¯¯¯B)

f=¯¯¯AB+C(¯¯¯A(B+¯¯¯B)+A¯¯¯B+B)

Distribute the ¯¯¯A

f=¯¯¯AB+C(¯¯¯AB+¯¯¯A¯¯¯B+A¯¯¯B+B)

AND the B term with 1 in the form of (A+¯¯¯A)

f=¯¯¯AB+C(¯¯¯AB+¯¯¯A¯¯¯B+A¯¯¯B+(A+¯¯¯A)B)

Distribute the B:

f=¯¯¯AB+C(¯¯¯AB+¯¯¯A¯¯¯B+A¯¯¯B+AB+¯¯¯AB)

The two terms in red are duplicated, therefore one can be eliminated:

f=¯¯¯AB+C(¯¯¯AB+¯¯¯A¯¯¯B+A¯¯¯B+AB+¯¯¯AB)

f=¯¯¯AB+C(¯¯¯A¯¯¯B+A¯¯¯B+AB+¯¯¯AB)

The 4 terms are all of the possible case of A AND B, therefore, it is 1:

f=¯¯¯A