bool arith

[stocktroll]

symbols:
b' = not b
a.b = a and b
a + b = a or b


This is what i thought should be a simple problem got me stumped since I cant simplify to the given answer...

Simplify the boolean expression:
a.b’.c’.d + a.b’.c.d + a.b.c’.d + a.b.c.d’ + a.b.c.d

the most i can simplify is a.b'.d + a.c'.d + a.b.c
but the answer given is: a.d + a.b.c

anyone see how this is so?

[losCHUNK]

you cant add letters you stupid cawk

[Guest]

what class do you learn that in?

[MKJ]

why, thinking of enrolling after the answer was given in this thread ?

[jester!]

what class do you learn that in?

I learned the simple stuff with that and Venn diagrams in Logic.

[stocktroll]

lol the hell i remember academic crap as i cruise controlled throughout college but this is what i suppose you should learn intro discrete math and probably in get some in intro course about computer architecture.... its the basis of taking 1's and 0's in circuits and doing all the cool crap computers can do

btw any of you noobs got a clue if they can simplify to the given answer?

[Guest]

Sorry, I haven't done that branch of mathematics.

[Pext]

rather easy. maybe you should learn what boolean arithmetic actually is:

http://mathworld.wolfram.com/BooleanArithmetic.html

i'll leave the . out. it's just like you can leave them out whith multiplication.

btw: 1+1 = 0
b' = 1+b = b-1 or 1-b... whatever.

~1

a(1+b)(1+c)d
= a(1+b)d + a(1+b)cd
= ad + abd + acd + abcd

~2

a(1+b)cd
= acd + abdc

~3

ab(1+c)d
= abd + abcd

~4

abc(1-d)
= abc + abcd


~adding it all together:

ad + abd + acd + abcd
+ acd + abdc
+ abd + abcd
+abc + abcd
= ad + abc


q.e.d.

[Guest]

owned