how to solve the following equation

[faiza nasir]

IF A=0 in logic expression Z= [A+EF+BC+D][A+DE+BC+DE],then
A) Z=0
B) Z=1
C) Z=BC
D) Z=BC
Where the bold characters are complements.

plz solve with complete steps.

 

[albbg]

There are no bold characters in your expression. I think it's better to use "!" in front of the character you want to complement; f.i.; !A means "not A".
Of course with BC I think you mean "B and C". Is it correct ? In this case !BC will mean "not B and C" while B!C will mean "B and not C"

 

[mathespbe]

IF A=0 in logic expression Z= [A+EF+BC+D][A+DE+BC+DE],then
A) Z=0
B) Z=1
C) Z=BC
D) Z=BC
Where the bold characters are complements.

plz solve with complete steps.

There are no bold characters in your expression...

 

[faiza nasir]

ok sorry for incorrect equation here's the correct equation

IF A=0 in logic expression Z= [A+EF+!BC+D][A+!D!E+!BC+!D!E],then
A) Z=0
B) Z=1
C) Z=BC
D) Z=BC

plz solve with complete steps.

 

[mathespbe]

Z= [A+EF+!BC+D][A+!D!E+!BC+!D!E]
if A = 0 ==> Z= [EF+!BC+D][!D!E+!BC+!D!E]
take !BC as common term,
Z = !BC[(EF+1+D)(!D!E+1+!D!E)]
since 1+A = A,
Z = !BC[(EF+D)(!D!E+!D!E)]
since A+A = A
Z = !BC[(EF+D)(!D!E)]
now
Z = !BC[EF!D!E + D!D!E]
since A!A = 0
Z = !BC[0+0]
==> Z = 0

 

[albbg]

you have first to expand the equation, like a normal expession remembering that:
x+x=x*x=x
x+!x=1
x*!x=0

using these rules and rearranging I obtained (hope I didn't do errors):

Z=a*(1+!be+!bc+ef+d)+!bc*(ef +!d!e+1+d)

so in the brackets there is always 1.

Z=a+!bc

then Z=1 if a=1 or !bc=1 that is b=0 and c=1
z=0 if a=0 and !bc=0
z=bc never
z=!bc if a=0

I'm from a mobile phone and is difficult to explain better. on saturday I'll have access to my PC.