I can't simplify this A’B’C + A’BC + A’BC’ + AB’C + ABC boolean expression to A'B+C I have to get this expression A’B’C + A’BC + A’BC’ + AB’C + ABC to A'B+C. I did this but I can't finish it, I don't know how to.
A’B’C + A’BC + A’BC’ + AB’C + ABC
A'B(C+C')+C(A'B'+AB'+AB)
A'B+C(A'B'+AB'+AB)
That's it, I don't know how to solve that.
 A: \begin{align}
E&=A'B+C(A'B'+AB'+AB)\\&=A'B+C((A'+A)B'+AB)\\ &=A'B+C(B'+AB)
\end{align}
For boolean expressions $+$ also distributes over product, by example:
$$
B'+AB = (B'+A)(B'+B)=B'+A
$$
thus
\begin{align}
E &=A'B+C(B'+A)\\ 
&=A'B+CB'+CA \\
&=(A'B+C)(A'B+B')+CA \\
&=(A'B+C)(A'+B')+CA\\
&=A'B+CA'+CB'+CA \\
&=A'B+C+CB' \\
&=A'B+C
\end{align}
A: Terms can be duplicated in boolean algebra, because or and and are idempotent: $X = X+X$ and $X = XX$.
$$\begin{align*}
p &= A’B’C + \underline {A’BC} + A’BC’ + AB’C + ABC\\
&= (\underline {A’BC} + A'BC') + (A'B'C + \underline {A’BC} + AB'C + ABC)\\
&= A'B(C+C') + (A'+A)(B'+B)C\\
&= A'B + C
\end{align*}$$
(Underlined terms were duplicated, no special meaning otherwise)
Also try using Karnaugh map which helps a lot.
A: There is this super-weird absorption laws in boolean algebra: $X+XY = X$ and $X(X+Y) = X$.
From your last line, as you "prematurely grouped" the $A'BC$ term into $A'B$, the absorption law can "recreate" an $A'BC$ out of $A'B$:
$$\begin{align*}
p &= A'B+C(A'B'+AB'+AB)\\
&= (A'B + A'BC) + C(A'B ' + AB' + AB)\\
&= A'B + C(A'B + A'B' + AB' + AB)\\
&= A'B + C(A' + A)(B'+B)\\
&= A'B + C
\end{align*}$$
Also try using Karnaugh map which helps a lot.
A: $$A'B'C'+A'BC'+A'BC+ABC=$$
$$=A'C(B'+B)+A'BC+ABC'   , \{B'+B=1\}$$
$$=A'C'+B(A'C+AC')$$
$$=A'C' +B$$
