Understanding Absorption Laws (Boolean Algebras)

[mathrookie]

TL;DR

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)

 

[Mark44]

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.

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)

 

[TeethWhitener]

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.

 

[Mark44]

Slight typo here, should be $a\wedge(\top\vee b)$

Right. I've fixed it in my post.