Proof that $B \land ( B \lor C) = B$?

Question

In my logic design exam today I was given this question:

Show that: $$ B \land ( B \lor C) = B $$

It's asking for a proof for this expression. Could someone please explain how such expression can be proven? I'm not that good at Boolean algebra but I believe that it's in the simplest form.

Answer 1

You could simply do it by a truth table. Or use the simple facts that $x\land y$ implies $x$ as well as $x$ implies $x\lor y$. Thus $B\land(\ldots)$ implies $B$ and $B$ implies $B\lor C$ and hence also implies $B\land (B\lor C)$, in summary $B\land(B\lor C)$ and $B$ are equivalent.

Answer 2

Distribute out expression, and you're left with B and B = B or B and C. so that should imply B.

$B \wedge (B \lor C) = (B\wedge B) \lor (B\wedge C )= B \lor (B\wedge C ) = B$

Answer 3

A truth table will show it

B    C    B or C    B and (B or C)

T    T      T            T
T    F      T            T
F    T      T            F
F    F      F            F 

Answer 4

$a*(a+b)=a*a+a*b=a+a*b=a*1+a*b=a*(1+b)=a$

Conjunction $x∧y$ behaves on $0$ and $1$ exactly as multiplication does for ordinary algebra: if either $x$ or $y$ is $0$ then $x∧y$ is $0$, but if both are $1$ then $x∧y$ is $1$.

Disjunction $x∨y$ works almost like addition, with $0∨0 = 0$ and $1∨0 = 1$ and $0∨1 = 1$. However there is a difference: $1∨1$ is not $2$ but $1$.