$(1)(2$ marks$)$ Prove DeMorgan$$s Laws using the method of truth table:

$($a$)($x $+$ y$)=$ x$$y$$;

$($b$)($xy$)=$ x$+$ y$$

$(2)(4$ marks$)$ Simplify the following Boolean functions using only postulates and

theorems of Boolean algebra on the handout distributed in the class. Make sure that your

final answer cannot be further simplified. For each step, you need to specify which

postulate or theorem you have used.

$($a$)$ xyz$+$ x$$yz $+$ xyz $+$ x$$yz$$

$($b$)($x $+$ y$)($x$+$ y$)$

$(3)(4$ marks$)$ Reduce the following Boolean functions to the indicated number of literals

using postulate and theorems of Boolean Algebra. Recall that a literal is either a variable

itself or a variable$$s negation.

$($a$)$ wxyz $+$ w$$xz $+$ wxy$$z $($to two literals$)$

$($b$)($xz$+$ y$)+$ y $+$ xz $+$ wy $($to three literals$)$

$(4)(3$ marks$)$ Find the truth table for the following Boolean function and then draw its

logic diagram using AND, OR and NOT gates.

F$($x$,$ y$,$ z$)=$ x $+$ y$+$ y$($x $+$ z$)$

$(5)(4$ marks$)$ Find the complement for each of the following Boolean functions and

simplify them.

$($a$)($A$$B $+$ CD$)$E $+$ E$$

$($b$)($x$+$ y $+$ z$)($x $+$ y$)($x $+$ z$)$

$(6)(2$ marks$)$ Express, in sum$-$of$-$minterms, the complement of the following functions.

$($a$)$ F$($A$,$ B$,$ C$,$ D$)=$ m$(2,9,10,12,14).$

$($b$)$ F$($x$,$ y$,$ z$)=$ M$(1,4,5,7)$

$(7)(2$ marks$)$ Convert each of the following to the other standard form, i$.$e$.,$ from sum of

minterms to product of maxterms or from product of maxterms to sum of minterms.

$($a$)$ F$($x$,$ y$,$ z$)=$ m$(2,4,7)$

$($b$)$ F$($A$,$ B$,$ C$,$ D$)=$ M$(0,1,3,4,7,11,12)$

$(8)(4$ marks$)$ Simplify the following Booleans function using k$-$maps.

$($a$)$ F$($x$,$ y$,$ z$)=$ m$(0,1,6,7)$

$($b$)$ F$($x$,$ y$,$ z$)=$ xyz $+$ x$$y$$z $+$ xy$$z$$

$(9)(4$ marks$)$ Simplify the following Boolean functions using k$-$maps.

$($a$)$ F$($w$,$ x$,$ y$,$ z$)=$ m$(1,4,5,6,8,13)$

$($b$)$ F$($A$,$ B$,$ C$,$ D$)=$ m$(0,2,3,5,6,7,8,10,13,15)$

$(10)(3$ marks$)$ Simplify the following Boolean function using k$-$maps.

F$($A$,$ B$,$ C$,$ D$)=$ C$$D$+$ A$$B$$C $+$ ABC$+$ AB$$C