how to prove $A+A'B+A'B'C+A'B'C'D=A+B+C+D$ Prove the above relationship by using the Boolean definition. I tried $A+A'B=A+B$, but end up with $A+B+A'B'(C+D)$, how can I go next?


I tried $A+A'B=A+B$

Apply this to $\,A+A'B'C=A+B'C\,$ then again to $\,B+B'C=B+C\,$:
$$
\begin{align}
A+A'B+A'B'C &= \color{red}{A+B}+A'B'C
\\ &= \color{red}{A}+B+\color{red}{B'C}
\\ &= A+\color{red}{B+C}
\end{align}
$$
Repeat the steps to prove the equality with all four terms.

[ EDIT ] $\;$ Another way also using OP's attempt, only working from the end backwards.
$$
\begin{align}
A+A'B+A'B'C+A'B'C'D &= A+A'B+A'B'(\color{blue}{C+C'D})
\\ &= A+A'B+A'B'(\color{blue}{C+D})
\\ &= A+A'\big(\color{red}{B+B'(C+D)}\big)
\\ &= A+A'\big(\color{red}{B+(C+D)}\big)
\\ &= \color{green}{A+A'(B+C+D)}
\\ &= \color{green}{A+(B+C+D)}
\\ &= A+B+C+D
\end{align}
$$
