Prove using mathematical proof:

$\displaystyle xyz + \overline{x}z = yz + \overline{x}z$

I can't figure how to begin with this. Help appreciated. thanks.

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- Sep 27th 2013, 08:28 AMlarry21Prove using mathematical proof
Prove using mathematical proof:

$\displaystyle xyz + \overline{x}z = yz + \overline{x}z$

I can't figure how to begin with this. Help appreciated. thanks. - Sep 27th 2013, 09:56 AMlarry21Re: Prove using mathematical proof
- Sep 27th 2013, 11:14 AMemakarovRe: Prove using mathematical proof
$\displaystyle xy+\bar{x} = xy+\bar{x}(y+1) = xy+\bar{x}y+\bar{x} = (x+\bar{x})y+\bar{x} = y+\bar{x}$.

- Sep 27th 2013, 05:45 PMlarry21Re: Prove using mathematical proof
- Sep 27th 2013, 07:23 PMtopsquarkRe: Prove using mathematical proof
- Sep 27th 2013, 09:12 PMlarry21Re: Prove using mathematical proof
Very sorry for being unclear in my question and causing such confusion. But I only require to prove the LHS of the equation, thus by using the boolean postulates, laws, theorems to collapse the LHS to the RHS. I've studying this equation for so long and have yet been able to achieve such a task. A complete solution is not required, I just need a starting point to guide me in the right direction. Thank you in advance for your help.

Edit:

$\displaystyle xyz + \overline{x}z = yz + \overline{x}z$

z(xy + x') = distributive law (factor)

z(x' + x)(x' + y) = distributive law (expand)

z(1)(x' + y) = complement

z(x'+y) = (anything ANDed by 1 equals itself)

zx' + zy = distributive (expand)

yz + x'z = commutative

I believe I've finally figured it out. Anyone care to critique? Thank you. - Sep 28th 2013, 06:41 AMemakarovRe: Prove using mathematical proof
As you can note, the key part of your solution is converting xy + x' to x' + y. The rest, i.e., multiplying by z and using distributivity, is trivial. The nontrivial part is what I showed in post #3. In fact, your transformation is shorter because you used distributivity of addition over multiplication, which allows changing xy + x' to (x' + x)(x' + y). So, good job!