How Can Algebraic Manipulation Prove the Consensus Property in Logic Circuits?

[shamieh]

Some background: I am in EE 280 Design of Logic Circuits.

Problem: Use algebraic manipulation to prove that xy +yz +x!z = xy + x!z. (Note that this is the consensus property which is: xy + yz + x!z = xy + x!z)

+ mean OR, ! mean NOT.

Please help! I am lost. I do have the rules near me (x AND 1 = x etc.. as well as the "Single Variable Theorems") If someone could walk me through solving this that would be great and I would be forever thankful.

 

[Ackbach]

Try this:
$$xy+\bar{x}z+yz=xy+\bar{x}z+(x+\bar{x})yz.$$

 

[shamieh]

Awesome, just figured it out. I solved and everything. I have a question tho. Where exactly does the rule (x + x!) come in play? I guess my question essentially is; how do I know that I can randomly put in a (x + x!) in the 2nd term. What is the property or rule that tells me I can do that legally? Sorry if this seems like a dumb question I'm just trying to understand what I'm actually doing versus just solving the equation.

Thanks again Ackbach,
-Sham(Ninja)

 

[Ackbach]

Well, $x+\bar{x}=\text{T}$, and $Tz=z$. So you can always multiply anything by $T$ and not change the truth value.

 

[blue_raver22]

Sure, I'd be happy to help you solve this problem! Let's start by breaking down the problem and understanding what the different symbols mean.

First, we have the variables x, y, and z. In this context, these variables represent individual inputs to a logic circuit. So, for example, x could represent an input that is either 0 or 1, and y could represent another input that is also either 0 or 1.

Next, we have the symbols + and !. In this context, + represents the OR operation, which means that the output will be 1 if either or both of the inputs are 1. The ! symbol represents the NOT operation, which means that the output will be the opposite of the input (if the input is 1, the output will be 0, and vice versa).

Now, let's look at the left side of the equation: xy + yz + x!z. We can break this down into three separate parts: xy, yz, and x!z. Each of these parts represents a different logical operation being performed on the inputs.

The first part, xy, represents the AND operation. This means that the output will be 1 only if both inputs (x and y) are 1. The second part, yz, also represents the AND operation, but with different inputs (y and z). Finally, the third part, x!z, represents the NAND operation, which is the opposite of AND. This means that the output will be 0 only if both inputs (x and z) are 1.

Now, let's look at the right side of the equation: xy + x!z. This is a simpler expression, with only two terms. However, we can see that it still represents the same logical operation as the left side: AND and NAND.

To prove that these two expressions are equivalent, we can use algebraic manipulation. Let's start by expanding the left side of the equation:

xy + yz + x!z
= xy + yz + x(!z) // using the fact that !z = (z)
= xy + yz + (xz) // using the fact that !z = (z)
= (xy + yz) + (xz) // using the associative property of addition
= (xy + yz) + (x!z) // using the fact that !z = (