# Understanding Absorption Laws (Boolean Algebras)

• I
• mathrookie
In summary: Thanks.In summary, the absorption law is obtained by using the distributive law in Boolean Algebras. This allows for the expression to be simplified to ##a∧(a∨b)## and ultimately to ##a##. Some people may refer to it as distribution law, but it is the same concept.
mathrookie
TL;DR Summary
I cannot apply distribution law
I can't understand how absorption law is obtained. I get following steps.##a∨(a∧𝑏) = (a∧⊤)∨(a∧𝑏)##
##=(a∨a)∧(a∨b)∧(⊤∨a)∧(⊤∨b)##
then,

I come up with ##=a∧(a∨b)∧⊤∧⊤## so ##=a∧(a∨b)##

But, I cannot get ##a∧(⊤∨𝑏)##, as shown on here, therefore ##a##.

Can you help me? I cannot obtain ##a∧(⊤∨𝑏)## Some people say in other answers in different questions, it is obtained by distribution law. However, what I got by this is the first equation.
[1]: https://proofwiki.org/wiki/Absorption_Laws_(Boolean_Algebras)

mathrookie said:
TL;DR Summary: I cannot apply distribution law

I can't understand how absorption law is obtained. I get following steps.##a∨(a∧𝑏) = (a∧⊤)∨(a∧𝑏)##
##=(a∨a)∧(a∨b)∧(⊤∨a)∧(⊤∨b)##
##a ∨ (a∧𝑏) = (a∧⊤)∨(a∧𝑏)##
##= a ∧ (T ∨ b) ## ∧ distributes over ∨
## = a ∧ T = a## T ∨ b = T
Edited to fix earlier typo.
mathrookie said:
then,

I come up with ##=a∧(a∨b)∧⊤∧⊤## so ##=a∧(a∨b)##

But, I cannot get
##a∧(⊤∨𝑏)##, as shown on here, therefore ##a##.

Can you help me? I cannot obtain
##a∧(⊤∨𝑏)## Some people say in other answers in different questions, it is obtained by distribution law. However, what I got by this is the first equation.
[1]: https://proofwiki.org/wiki/Absorption_Laws_(Boolean_Algebras)

Last edited:
Mark44 said:
##a ∨ (a∧𝑏) = (a∧⊤)∨(a∧𝑏)##
##= a ∧ (T ∧ b) ## ∧ distributes over ∨
## = a ∧ T = a## T ∨ b = T
Slight typo here, should be ##a\wedge(\top\vee b)##
OP, you can also use a truth table to see that the two expressions must be equal to a.

TeethWhitener said:
Slight typo here, should be ##a\wedge(\top\vee b)##
Right. I've fixed it in my post.

TeethWhitener

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