Understanding Absorption Laws (Boolean Algebras)

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)##

Your expression above doesn't help.
Follow the logic in your link to get this:
##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)

Mark44 said:

Your expression above doesn't help.
Follow the logic in your link to get this:
##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.