Prove that $(a\lor b)\land a\equiv a$ without using truth tables

Question

Please help me solve $(a\lor b)\land a\equiv a$ without using a truth table. Meaning that if there exists a choice between two things ($a$ or $b$) and one of them had to be chosen ($a$) then that is the same as simply $a$.

Answer 1

$ (a\vee b)\wedge a\equiv (a\vee b)\wedge (a\vee F)\equiv a\vee(b\wedge F)\equiv a\vee F\equiv a $

Answer 2

You might find it easier to prove that $a$ implies $(a\lor b)\land a$ and that $(a\lor b)\land a$ implies $a$. This means that they imply each other and are equivalent logically.

Answer 3

If you had to solve this using boolean algebra, then this would normally simply be one of the equivalence principles:

Absorption

$P \land (P \lor Q) = P$

$P \lor (P \land Q) = P$

... but since that would make the problem trivial, I assume you are not asked to do this using boolean algebra.

So, what's left? Maybe a formal proof? Many proof systems exist, with different sets of rules, so unless you tell us that you need to do a formal proof, an wat rules you have, I'll skip this method as well.

Instead, I'll just do a non-formal proof based on the semantics of the operators involved:

If $(a \lor b) \land a$ is true then both $a \lor b$ and $a$ are true, from whch it immediately follows that $a$ is true. So, $(a \lor b) \land a$ implies $a$ is true.

If $a$ is true, then $a \lor b$ will be true as well, since at least one of $a$ and $b$ is true. So, both $a$ and $a \lor b$ is true, and therefore $(a \lor b) \land a$ is true. So, $a$ implies $(a \lor ) \land a$.

Since the two statements imply each other, they are equivalent.