I want to prove the following logic equation in logic algebra
$$ a \bar b \bar e f + \bar a \bar b ef+ ac \bar d \bar e + \bar a c \bar d e+ \bar b \bar c f + \bar b d f = ac \bar d \bar e + \bar a c \bar d e+ \bar b \bar c f + \bar b d f $$
Last 4 terms in the LHS are equal to the RHS. According to this fact I tried to show that
$$a \bar b \bar e f + \bar a \bar b ef = 0$$
But I couldn't do that. Is there a particular method to solve this kind of equation ( terms in RHS are in the LHS)?
On
It is not correct to simply omit terms from both sides which overlap remaining terms.
You can simplify both sides of the equation to
$$\bar b \bar c f + \bar b d f + \bar a c \bar d e + a c \bar d \bar e$$
Another way would be to write truth-tables or to list the minterms for both sides and compare the six-literal minterms.
There are more elaborate methods to compare complex functions. Ask your favorite search engine for Equivalence Checking.
Your attempts to show that $a\bar{b}\bar{e}f + \bar{a}\bar{b}ef = 0$ are vain since this is not true (take, for example, $a = f = 1, b = e = 0$). More generally, every non-trivial sum of products is not $0$, you just need to choose the values of variables properly to satisfy one particular summand.
To eliminate these summands you can use the absorption law $A + AB = A$. But since these summands are not absorbed by any other summand we need to add extra variables to them using the complementation law $A = aA + \bar{a}A$.
We have the following chains of transformations for each of the summands: \begin{align} a\bar{b}\bar{e}f = a\bar{b}c\bar{e}f + a\bar{b}\bar{c}\bar{e}f = a\bar{b}cd\bar{e}f + a\bar{b}c\bar{d}\bar{e}f + a\bar{b}\bar{c}\bar{e}f.\\ \bar{a}\bar{b}ef = \bar{a}\bar{b}cef + \bar{a}\bar{b}\bar{c}ef = \bar{a}\bar{b}cdef + \bar{a}\bar{b}c\bar{d}ef + \bar{a}\bar{b}\bar{c}ef. \end{align}
Now observe that each of these six ($3 + 3$) summands is absorbed by one of the last four terms in the LHS of the original equation (for example, $a\bar{b}cd\bar{e}f$ is absorbed by $\bar{b}df$ and so on). So both $a\bar{b}\bar{e}f$ and $\bar{a}\bar{b}ef$ are absorbed producing that the LHS equals to the RHS.