Proving logic equation in logic algebra I'm trying to prove the following logic equations are equal and am having trouble.

$$\begin{align}
a\overline b\overline ef + \overline a\overline bef + ac\overline d\overline e + \overline ac\overline de + \overline b\overline cf + \overline pdf &= ac\overline d\overline e + \overline ac\overline de + \overline b\overline cf + \overline bdf\\
\overline a &= \neg a
\end{align}$$

I am pretty new to this so if someone can give me the simplest way possible it would be much appreciated. I have been trying to do it by simplification but cant get anywhere. I then tried using truth tables but I am obviously doing it wrong. So if someone could give a detailed and easy to understand way of how this can be solved with truth table as I said it would be much appreciated.
Is it possible to do this with K-maps?
 A: It is not difficult at all
to compare boolean equations like this via truth table,
if we have nothing more than any spreadsheet (Gnumeric, Excel) at hand.
Note that the boolean equation can be easily 
transformed into arithmetic one, e.g.:
\begin{align}
ab'e'f+a'b'ef+acde' 
&=
\max(a(1-b)(1-e)f,(1-a)(1-b)ef,acd(1-e))
\end{align}
First define the alphabet (the number of different variables) and its size, $n$.
For the given example we have $n=6$ variables $a,b,c,d,e,f$, 
hence the truth table is $2^n=64$ lines long.
All possible input combinations can be created 
following an easy-to remember staircase pattern.
Start with the entitling the columns with the symbols 
$a,b,c,d,e,f$ and their negation $a',b',c',d',e',f'$.
The $f$ variable follows alternating pattern $01010101...$,
every other doubles the number of $0$s and $1$s: 
$e\to 001100110011...$,
$d\to 00001111.00001111...$,
the first column has $2^{n-1}$ zeros
followed by $2^{n-1}$ ones:


Next, label columns with the components of the formula (the arguments of $\max$),
type according formulas in the top row and replicate the formulas downto the bottom:
 
Then set up a column that calculates a maximum of the items $=\max(...)$ 
and the truth table is finished for the left hand side of the equation:

Similarly handle the right hand side and compare the two columns.
A: You have too many literals. In this case you have 6.Your truth table should have 2^6 rows (interpretations). To show it's false. The given logical formula should be  false for all 2^6 interpretations which is extremely tedious to work by hand. Instead, you use truth tree method to show its falsity. The video explains how to do it Truth Tree. You have to use inference rules to do it.
A: $$ a \bar b \bar e f + \bar a \bar b ef + ac \bar d \bar e + \bar a c \bar d e + \bar b \bar c f + \bar b df = ac \bar d \bar e + \bar a c \bar d e + \bar b \bar c f + \bar b df$$
$$a \bar b \bar e f + \bar a \bar b ef + ac \bar d \bar e + \bar a c \bar d e + \bar b \bar c f + \bar b df$$ 
Look for common terms and duplicate as needed. 
Once you check this you find that each term has two different terms.  So we must expand to minimize. 
The key here is to realize the answer is given. The last four terms are the same in both expressions. It means the first two terms are redundant.
Leave the last four terms as they are. Expand first two terms and see if they can be absorbed by the others.
Easier if you expand first, then second. I will do first the second I leave to you.  
Keys:
$ X + 1 = 1$ Annulment;
$X + \bar X = 1$ Complement;
$$a \bar b \bar e f  ( \bar c \bar d + \bar c d + c \bar d + cd) + \bar a \bar b ef + ac \bar d \bar e + \bar a c \bar d e + \bar b \bar c f + \bar b df $$
$$a \bar b \bar c \bar d \bar e f + a \bar b \bar c d \bar e f + a \bar b c \bar d \bar e f + a \bar b cd \bar e f + \bar a \bar b ef + ac \bar d \bar e + \bar a c \bar d e + \bar b \bar c f + \bar b df $$
Look for common terms in first four terms with last 4.
$$(a \bar b \bar c \bar d \bar e f + a \bar b \bar c d \bar e f + \bar b \bar c f) + (a \bar b c \bar d \bar e f + ac \bar d \bar e ) + (a \bar b cd \bar e f + \bar b df)+ \bar a \bar b ef + \bar a c \bar d e $$
$$\bar b \bar c f (a \bar d \bar e + ad \bar e + 1) + ac \bar d \bar e ( \bar b f + 1) + \bar b df (ac \bar e + 1) + \bar a \bar b ef + \bar a c \bar d e$$
$$\bar b \bar c f + ac \bar d \bar e + \bar b df + \bar a \bar b ef + \bar a c \bar d e$$
Rearranging
$$\bar a \bar b ef + ac \bar d \bar e + \bar a c \bar d e + \bar b \bar c f + \bar b df$$
6 terms have become 5. Now expand the 1st term the same way.
Boolean Laws
