Name: Zahid
Student ID:
Maqbool
6572017221
a. Prove the validity of the logic equation (A+C) (AB+C’)=AB+AC’ by using the truth table and
waveform simulation.
Solution:
Truth table:
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
schematic diagrams:
AB+AC’
VHDL Code:
C’
1
0
1
0
1
0
1
0
A+C
0
1
0
1
1
1
1
1
AB
0
0
0
0
0
0
1
1
AB+C’
1
0
1
0
1
0
1
1
(AB+C’)+(A+C)
0
0
0
0
1
0
1
1
AC’
0
0
0
0
1
0
1
0
AB+AC’
0
0
0
0
1
0
1
1
Waveform:
Given Equation: (AB+C’) +(A+C)
Distribute
(A+C) (AB+C') = A(AB+C') + C(AB+C')
= AAB + AC' + CAB + CC'
= AB + AC' +0
So, the simplified expression is AB + AC'
The equation (A+C) (AB+C') = AB+AC' simplifies to AB + AC' + C using Boolean theorems.
Question 2
Prove the validity of the logic equation ABC +AB’(A’C’)’=A(C+B’) by using the truth table and
waveform simulation.
Truth table:
A
0
0
0
0
1
1
1
1
B
0
0
1
1
0
0
1
1
C
0
1
0
1
0
1
0
1
A’
1
1
1
1
0
0
0
0
schematic diagrams:
B’
1
1
0
0
1
1
0
0
C’
1
0
1
0
1
0
1
0
ABC
0
0
0
0
0
0
0
1
AB’
0
0
0
0
1
1
0
0
A’C’
1
0
1
0
0
0
0
0
(A’C’)’
0
1
0
1
1
1
1
1
(A’B’)’AB’
0
0
0
0
1
1
0
0
(A’C’)’AB’+ABC
0
0
0
0
1
1
0
1
C+B’
1
1
0
1
1
1
0
1
A(C+B’)
0
0
0
0
1
1
0
1
VHDL code:
Waveform:
Boolean theorem
Given Equation: ABC + AB' (A'C')' = A(C + B')
De Morgan's Theorem: (A'B')' = A + B
A Morgan's theorem to the term (A'C')':
(A'C')' = A + C
Distributive Law: AB + AC = A(B + C)
Apply the distributive law to the term A(C + B'):
A(C + B') = AC + AB'
Commutative Law: AB = BA
Rearrange the term AB' to B'A:
AB' = B'A
Now, let's substitute the simplified terms back into the original equation:
Original Equation: ABC + AB' (A'C')' = A(C + B')
Substitute De Morgan's theorem and the rearranged term:
ABC + B'A (A + C) = AC + AB'
Finally, using the distributive law again:
A(B'C) + B'A = A(C + B')
Which is the same as the right-hand side of the original equation A(C + B').
So, we've successfully proven that ABC + AB' (A'C')' is indeed equal to A(C + B') using Boolean theorems.