A postulate of boolean algebra: x+(y.z) = (x+y).(x+z)

In summary, the equation x+(y.z) = (x+y).(x+z) can be proven using proof by exhaustion in Boolean algebra. The statement x OR (y AND z) = (x OR y) AND (x OR z) is equivalent to the RHS, and in order for the RHS to be true, x must be true or both y and z must be true. This can also be proved by starting from the RHS and expanding it to equal x + (y.z). Additionally, there are no other expressions equivalent in value to the quoted expressions. The phrase "teaching granny to suck eggs" originated from the 18th century and means explaining something to someone who already knows it.
  • #1
jackson6612
334
1
x+(y.z) = (x+y).(x+z)

I don't understand how one gets the RHS. Could you please it? There must be some reasoning involved. Please help me. Thanks.
 
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  • #2
Isn't that an axiom?
 
  • #3
Yes, it's distributivity.

x OR (y AND z) = (x OR y) AND (x OR z).

If the RHS is true, x must be true, or both y and z must be true, which is the statement of the LHS.
 
  • #4
jackson6612 said:
x+(y.z) = (x+y).(x+z)

I don't understand how one gets the RHS. Could you please it? There must be some reasoning involved. Please help me. Thanks.

One way to prove it is by using proof by exhaustion. If we were doing normal algebra we couldn't do this but because the state space of Boolean algebra is small in this example (8 possibilities) its easy to prove it in this way.
 
  • #5
micromass said:
Isn't that an axiom?

Hi Micro

Isn't a postulate a same thing as an axiom? Perhaps, in mathematical terms there is a distinction but M-W does take them equivalent in some respects. Please let me know.

Thank you, everyone, for all the help. Makehc, your post was quite helpful.
 
  • #6
Starting from RHS:

(x+y).(x+z)=x.x + x.z + x.y + y.z
=x + x.z + x.y + y.z
=x.(1 + z + y) + y.z
=x + (y.z)

Apologies if this is teaching granny to suck eggs, or indeed if it's wrong. This is how I would visualise it.
 
  • #7
jackson6612 said:
Hi Micro

Isn't a postulate a same thing as an axiom? Perhaps, in mathematical terms there is a distinction but M-W does take them equivalent in some respects. Please let me know.

Thank you, everyone, for all the help. Makehc, your post was quite helpful.

Yes, a postulate is the exact same thing as an axiom (although there was a distinction in the past). But I always use the word axiom. Sorry for the confusion!
 
  • #8
micromass said:
Yes, a postulate is the exact same thing as an axiom (although there was a distinction in the past). But I always use the word axiom. Sorry for the confusion!

No worries!
 
  • #9
cobalt124 said:
Starting from RHS:

(x+y).(x+z)=x.x + x.z + x.y + y.z
=x + x.z + x.y + y.z
=x.(1 + z + y) + y.z
=x + (y.z)

Apologies if this is teaching granny to suck eggs, or indeed if it's wrong. This is how I would visualise it.

Cobalt: Thank you. BTW, I'm no granny, just a layman!
 
Last edited:
  • #10
x+(y.z) = (x+y).(x+z)

Isn't there any other expression(s) which could be equivalent in value to the quoted expressions? Something like (x+y).(x+z).(z.z)? I know it's wrong; I'm simply trying to explain what I have asked you for. Thanks.
 
  • #11
Ha Ha, and I managed to insult you as well! It's teaching YOUR granny to suck eggs. I should find out where the expression came from before I use it.
 

1. What is a postulate of boolean algebra?

A postulate of boolean algebra is a fundamental rule or principle that serves as a basis for the manipulation and simplification of boolean expressions. It helps in understanding the properties of boolean operations and their relationships.

2. What does the postulate x+(y.z) = (x+y).(x+z) mean?

This postulate states that the boolean operation "or" between a variable x and the product of two other variables y and z, is equivalent to the "and" operation between x and the sum of y and z. In other words, it shows the relationship between the "or" and "and" operations in boolean algebra.

3. Why is this postulate important in boolean algebra?

This postulate is important because it helps in simplifying boolean expressions and allows for the conversion of complex expressions into simpler forms. It also provides a basis for developing more advanced rules and theorems in boolean algebra.

4. How is the postulate x+(y.z) = (x+y).(x+z) applied in problem-solving?

This postulate is applied in problem-solving by allowing us to manipulate boolean expressions and simplify them using the given relationship between the "or" and "and" operations. It helps in finding the most simplified form of a boolean expression, which is useful in various applications such as digital logic design and computer programming.

5. Are there any other postulates in boolean algebra?

Yes, there are several other postulates and principles in boolean algebra, such as the commutative, associative, and distributive laws, De Morgan's laws, and the identity and complement laws. These postulates, along with the one mentioned above, form the basis of boolean algebra and are used extensively in solving problems and proving theorems.

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