# Proof on discrete maths, help :)

• Oct 29th 2009, 03:46 PM
primeimplicant
Proof on discrete maths, help :)
I 'm trying to proove that :

(a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c)

i ve put it on truth table and its working, but i have to do it actually without truth table, so i tryed boolean algebra.

I think that i am making a fatal mistake but i dont know where it is. So i started from this with boolean algebra --> (a OR b OR c)AND(-a OR -b OR -c) trying to reach 1st part, but actually this is what i achived :

aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc' =
ab' + ac' + ba' +bc' + ca' +cb' =
(a XOR c) + (b XOR c) + (a XOR b)

+ = logical OR
multiplies = logical AND

• Oct 30th 2009, 06:36 AM
HallsofIvy
Quote:

Originally Posted by primeimplicant
I 'm trying to proove that :

(a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c)

i ve put it on truth table and its working, but i have to do it actually without truth table, so i tryed boolean algebra.

I think that i am making a fatal mistake but i dont know where it is. So i started from this with boolean algebra --> (a OR b OR c)AND(-a OR -b OR -c) trying to reach 1st part, but actually this is what i achived :

aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc' =
ab' + ac' + ba' +bc' + ca' +cb' =
(a XOR c) + (b XOR c) + (a XOR b)

+ = logical OR
multiplies = logical AND

You do understand, don't you, that "aa' " means "a is true and a is false"? In "boolean" aa'= bb'= cc'= 0.

The left side of (a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c) reduces to ab'+ bc'+ a'c and the right side to (a+ b+ c)(a'+ b'+ c').

Now, apply booean algebra to those.
• Oct 30th 2009, 07:21 AM
primeimplicant
Quote:

Originally Posted by HallsofIvy
You do understand, don't you, that "aa' " means "a is true and a is false"? In "boolean" aa'= bb'= cc'= 0.

Ehm, apparently i understand that, thats why from this point :
aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc'

i m going here :

ab' + ac' + ba' +bc' + ca' +cb' (i erase apparently aa', bb', cc' )

BUT in boolean algebra for example YZ' + ZY' = Y XOR Z ,

and the above statement --> ab' + ac' + ba' +bc' + ca' +cb'

equals (a XOR c) + (b XOR c) + (a XOR b) ,

I cant exactly understand where should i apply boolean, left side of equation or right side?

I am supposed to start from one side and result the other side.

I choose to start from right side but i didnt result left side .

What am i doing sooo wrong ? I think that i apply really bad boolean on right side, but where exactly?

• Oct 30th 2009, 03:18 PM
oldguynewstudent
Quote:

Originally Posted by primeimplicant
I 'm trying to proove that :

(a AND -b) OR (b AND -c) OR (c AND -a) = (a OR b OR c) AND (-a OR -b OR -c)

i ve put it on truth table and its working, but i have to do it actually without truth table, so i tryed boolean algebra.

I think that i am making a fatal mistake but i dont know where it is. So i started from this with boolean algebra --> (a OR b OR c)AND(-a OR -b OR -c) trying to reach 1st part, but actually this is what i achived :

aa' + ab' + ac' + ba' +bb' +bc' + ca' + cb' + cc' =
ab' + ac' + ba' +bc' + ca' +cb' =
(a XOR c) + (b XOR c) + (a XOR b)

+ = logical OR
multiplies = logical AND

(a $\wedge$ -b) $\vee$ (b $\wedge$ -c) $\vee$ (c $\wedge$-a) =
(a $\vee$ b $\vee$ c) $\wedge$ (a $\vee$ -c $\vee$ -a) $\wedge$ (-b $\vee$ b $\vee$ c) $\wedge$ (-b $\vee$ -c $\vee$ -a) Distributive property
= (a $\vee$ b $\vee$ c) $\wedge$ T $\wedge$ T $\wedge$ (-b $\vee$ -c $\vee$ -a) Negation and Domination Laws
= (a $\vee$ b $\vee$ c) $\wedge$ (-a $\vee$ -b $\vee$ -c) Domination and Commutative Laws