Logic Identity for $\mathcal{(A'B)}$+$\mathcal{(AB')}$

Question

I need to simplify a long boolean expression and I am stuck with this expression $$\mathcal{(A'B)}\text{+}\mathcal{(AB')}$$ Can I simplify that to $\mathcal{(AB)'}\large\text{ ?} $
And if so, what principle does that use$\large\text{ ?}$

Answer 1

No, both of your expressions are not equivalent. We can show this by Truth Table.

Since, $2$ variables are there, and both of these variables can assume $2$ values $[0/1]$.
Therefore, we can have $4$ possible combinations.

In general, if $n$ variables are there, we can have $2^n$ possible combinations.

Now, TRUTH TABLE

$\mathcal{A}$	$\mathcal{B}$	$\mathcal{A'}$	$\mathcal{B'}$	$\mathcal{A'B}$+$\mathcal{AB'}$	$\mathcal{AB}$	$\mathcal{(AB)'}$
$0$	$0$	$1$	$1$	$\LARGE{0}$	$0$	$\LARGE{1}$
$0$	$1$	$1$	$0$	$\LARGE{1}$	$0$	$\LARGE{1}$
$1$	$0$	$0$	$1$	$\LARGE{1}$	$0$	$\LARGE{1}$
$1$	$1$	$0$	$0$	$\LARGE{0}$	$1$	$\LARGE{0}$

As it is clear from table, both the columns are not same for all possible combinations.

Hence, the expressions are not equivalent.