Sum-of-products, simplication
F = /ABC + /AB/C + ABC + AB/C + /A/BC + /A/B/C

\$ F =  \bar{A}BC +\bar{A}B\bar{C}+ABC+AB\bar{C}+\bar{A}\bar{B}C+\bar{A}\bar{B}\bar{C} \$

We can plot this as a Karnaugh map

$$
\begin{array}{c|cccc}
  \text{}&amp;\bar{B}&amp;B&amp;B&amp;\bar{B}\\
  \hline
  \bar{A}&amp;1&amp;1&amp;1&amp;1 \\
  A&amp;0&amp;1&amp;1&amp;0 \\
  \hline
  \text{}&amp;\bar{C}&amp;\bar{C}&amp;C&amp;C
\end{array}
$$

Which by inspection we can see simplifies to \$B+\bar{A}\$

For completeness you should plot the truth table for both to confirm they are identical.



For an algebraic approach using DeMorgan

We can start by rearranging to collect all the \$ \bar{A} \$  terms.

\$ F = \bar{A}\bar{B}\bar{C}+\bar{A}\bar{B}C+\bar{A}B\bar{C}+\bar{A}BC+AB\bar{C} +ABC\$ 

Giving

\$F =\bar{A}\cdot(\bar{B}\bar{C}+\bar{B}C+B\bar{C}+BC)+AB\bar{C}+ABC \$

Now collecting on \$ \bar{B}\$ and \$ B \$

\$ F=\bar{A}\cdot(\bar{B}\cdot(\bar{C}+C)+B\cdot(\bar{C}+C))+AB\cdot{(\bar{C}+C)}\$

Now since \$ \bar{C} +C =1 \$ we can simplify further.

\$ F=\bar{A}\cdot(\bar{B}+B) +AB\$

and since \$ \bar{B} +B =1 \$ we get.

\$ F = \bar{A}+AB \$

Since \$ \bar{A} \$ does not care about \$B\$ we can see that we could add \$ \bar{A}B \$ without changing the result 

since \$ \bar{A} = \bar{A}\cdot(\bar{B}+B) = \bar{A} \bar{B} + \bar{A}B \$ = \$ \bar{A} + \bar{A}\bar{B} = \bar{A} + \bar{A}B \$

due to the property of idempotence

Giving.
`
\$F = \bar{A} +\bar{A}B+AB = \bar{A}+B\cdot(\bar{A}+A)=\bar{A}+B\$