Prove the following holds for all Boolean algebras?

Question

For all x,y: (x ∨ y) ∧ x = x

This is my attempt, I am just getting back to the question:

(x ∨ y) ∧ x

= x ∧ (x ∨ y)

= (x ∧ x) ∨ (x ∧ y)

= x ∨ (x ∧ y)

= (x ∨ x) ∧ (x ∨ y)

= x ∧ (x ∨ y)

So where am I going wrong here?

Answer 1

You are not "going wrong"; $x \land (x \lor y)$ follows. Its truth table has the same result as $x$.From line 4 above, we have $$\neg(\neg x) \lor (x \land y)$$

which is $\neg x \implies (x \land y)$. Applying the absorption rule as in the comment gives $$\neg x \implies (\neg x \land x \land y) $$

This is $x \lor F$ or $x$, as required.