Question

how to simplify:
\[(q\vee\neg r\vee p)\wedge(q\vee\neg r)\]

Answer 1

The easiest way is to use a truth table.

Answer 2

i want to use boolean algebra

Answer 3

from: (((p or q) implies r) or ((p or r) implies q)) implies (r implies q) i got posted thing and i can't simplify it more and it took like 10+ steps just to get that which i posted here and it's not final answer :/

Answer 4

are they the same?

Answer 5

yeah

Answer 6

wolframalpha says so

Answer 7

yeah you got correct answer

Answer 8

but it takes forever to simplify maybe im doing something wrong :/

Answer 9

Let me write it down then.

Answer 10

and still i didn't get simplified answer

Answer 11

if you don't have what to do you can write from this one
(((p or q) implies r) or ((p or r) implies q)) implies (r implies q)
lol

Answer 12

I will use + to indicate or, and . to indicate and.
\[(q+r'+p)(q+r')=q+(qr'+r'q)+r'+pq+pr'\]
\[=(q+qr')+(r'+pr')+pq=q(1+r')+r'(1+p)+pq\]
\[=q+r'+pq=r'+q(1+p)=r'+q.\]

Answer 13

(q+r'+p)(q+r')=q+(qr'+r'q)+r'+pq+pr' what rule did you used here

Answer 14

Few things you need to know:
1) p and p=pp=p
2) p or p=p+p=p
3) p or 1=p+1=1
4) p and 0=p.0=0

Answer 15

The rules I just wrote above.

Answer 16

have you just "multiplied" like in algebra?

Answer 17

I don't think the word "multiply" is accurate here, but yeah distribution is applicable.

Answer 18

@Tomas.A - this might help you: http://www.ee.surrey.ac.uk/Projects/Labview/boolalgebra/

Answer 19

i have seen distribution only when
(a and b) or c= (a or c) and (b or c)
(a or b) and c=(a and c) or (b and c)

Answer 20

i know how to do boolean algebra but when i get ultra large expression i always fail

Answer 21

It's the same thing but with more statements.

Answer 22

ok
so to simplify this
(((p or q) implies r) or ((p or r) implies q)) implies (r implies q)
it takes 15 steps and maybe half hour does it mean that i suck so hard that i create more complicated expressions

Answer 23

This is confusing! Could you rewrite it using latex?

Answer 24

http://www.wolframalpha.com/input/?i=%28%28%28p+or+q%29+implies+r%29+or+%28%28p+or+r%29+implies+q%29%29+implies+%28r+implies+q%29

Answer 25

it's the same just original question for those who start reading from here

Answer 26

How did you solve it?

Answer 27

i dont want to spend hour writing with latex everything i wrote on paper :D

Answer 28

I just meant did you solve it algebraically or with a truth table?

Answer 29

If I were you I would solve it by building a truth table.

Answer 30

algebraically, are you sure it would be easier with truth table?

Answer 31

ok i tried it again with algebra (from the original expression) but this time I noticed this rule:
(a and b) or a = a
and it simplified quite fast but took 9 steps anyway

Answer 32

Well, I used this rule above, but the notations I used confused you probably. I wrote \(ab+a=a(b+1)=a\).

Answer 33

yeah but this time i haven't even got that expression which u solved because i simplified earlier