Simplify $AC'+A'C+BCD'=AC'+A'C+ABD'$ How to prove that $$AC'+A'C+BCD'=AC'+A'C+ABD'$$ 
approch: a way to demonstrate is expressed in its canonical form.
Any hint would be appreciated.

In 
$$AC'+A'C+BCD'=AC'+A'C+ABD'$$ 
If any term out of $AC'$ and $A'C$ becomes $1$, then, both the LHS and RHS will become $1$ and the equality holds. 
So, the case left out is when both $AC'$ and $A'C$ are $0$. Then, suppose $A=1$. Then, $C'=0\Rightarrow C=1$. Similarly, if $A=0$, we have $A'=1$ which implies $C=0$.
Thus, $A$ and $C$ must have the same truth value in this case. Thus,
$BCD'$ and $ABD'$ will also have the same truth value. We can thus conclude that the inequality holds in all the cases.
