Simplify $AC'+A'C+BCD'=AC'+A'C+ABD'$

Question

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.

Answer 1

If $A=C$, then it is true by substitution.

If $A=\neg C$, then either $AC'$ is true or $A'C$ is true, making both sides true as well.

By the Law of the Excluded Middle, this is proven.

Answer 2

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.