Prove using Truth table:

a$)(A+B)(A^{\text{'}}+C)=AC+A^{\text{'}}B$

b$)(A+B)(A^{\text{'}}+C)(B+C)=(A+B)(A^{\text{'}}+C)$

Write down the simplified form of each expression, if you cannot simplify then explain why you

cannot. Also write down what theorem for simplification $($unite$/$absorption$/$elimination$\_$ is being

used

a$)A^{\text{'}}BC+A^{\text{'}}B{C}^{\text{'}}=?$

b$)BC+$ABCD$=?$

c$)AB+A{B}^{\text{'}}C=?$

d$)(A+B)(A+B+C^{\text{'}})=?$

e$)$ ABC$+A^{\text{'}}B{C}^{\text{'}}=?$

Given the following logic expression.

$F=A{B}^{\text{'}}C+$ABCD$+$ABC$+A^{\text{'}}CD$

a$)$ Draw the circuit using NOT, AND and OR gates. How many gates are needed?

b$)$ Now simplify the above expression $($you should be able to simplify to two product terms$).$

Then using the simplified form of the equation, draw the circuit again. How many gates are

needed now. Calculate percentage saving by computing the following:

$\%$ saving $=($number of gates needed before simplification $-$ number of gates needed after

simplification$)$ number of gates needed before simplification$)$

Convert the following SOP equation into a POS equation:

$F=A{B}^{\text{'}}+BCD+A{D}^{\text{'}}$

Simplify if you can.

Convert the following POS expression into SOP $($Try using theorems and simplify when possible$)$

$F=(A+B^{\text{'}})(A+B^{\text{'}}+C+D)(B^{\text{'}}+C+D)(B^{\text{'}}+C+D^{\text{'}})$