Boolean Algebra - Why is the result 1?

Question

Given:

= !(A * (!B + C)) + !(!B * !C)

= !A + (B * !C) + !B + C

Where: ! = NOT + = OR * = AND

I'm having some trouble to why !A + (B * !C) + !B + C simplifies to 1? Can someone shed some light on this please? It would be very much appreciated.

Answer 1

Use the distributive property to obtain an "or clause" consisting of $(B\; + \;!B + \cdots = 1)*(!C\;+ \;C + \cdots = 1)$

$$\begin{align} \lnot A + (B * \lnot C) + (\lnot B + C) & = \lnot A + (B + \lnot B + C)*(\lnot C + \lnot B + C)\\ \\ & = \lnot A + (1 + C)*(1+ \lnot B) \\ \\ &= \lnot A + 1*1 \\ \\ & = \lnot A + 1 = 1\end{align}$$

That is, $B \lor \lnot B$ is always true, as is $C \lor \lnot C$, and since the clause in which they occur one clause in a disjunction of clauses, the entire statement is there by true, since $P + (1)(1) = P + 1 = 1$, where $P$ can be any statements whatsoever.

Answer 2

If you draw the truth table, the result will be more obvious to you. Notice the three terms !A, !B and C. These account for 7 of the 8 possible states for (A,B,C). The only remaining state that's not accounted for is (A,B,!C). This is where the fourth term B * !C comes in.

Also, if you prefer seeing it worked out in Boolean Algebra:

I am going to be using ($+, \times, !$) as the operations.

$$!A + !B + (B\times!C) + C \\=\ !A + !B + !C + C \ \ (\ \because \ !B + B\times !C =\ !B+!C \ ) \\=\ !A + !B + 1 \\=\ 1 \ (\ \because X + 1 = 1 \ )$$