I need to simplify a long boolean expression and I am stuck with this expression
$$\mathcal{(A'B)}\text{+}\mathcal{(AB')}$$
Can I simplify that to $\mathcal{(AB)'}\large\text{ ?} $
And if so, what principle does that use$\large\text{ ?}$
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4$\begingroup$ No guy, you can't $\endgroup$– Mauro ALLEGRANZACommented Mar 14, 2021 at 11:12
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2$\begingroup$ That is actually the xor gate, it can be written as A $\oplus$ B $\endgroup$– Vinanth S BharadwajCommented Mar 14, 2021 at 11:29
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$\begingroup$ xor gate, but does that mean i can't simplify it more? $\endgroup$– Ross HendersonCommented Mar 14, 2021 at 11:33
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1$\begingroup$ How could it get simpler than "A (some operation) B"? anything else would have $0$ operators and so would be just A or just B, neither of which is same as what you start with. $\endgroup$– coffeemathCommented Mar 14, 2021 at 11:48
1 Answer
No, both of your expressions are not equivalent. We can show this by Truth Table.
Since, $2$ variables are there, and both of these variables can assume $2$ values $[0/1]$.
Therefore, we can have $4$ possible combinations.
In general, if $n$ variables are there, we can have $2^n$ possible combinations.
Now, TRUTH TABLE
$\mathcal{A}$ | $\mathcal{B}$ | $\mathcal{A'}$ | $\mathcal{B'}$ | $\mathcal{A'B}$+$\mathcal{AB'}$ | $\mathcal{AB}$ | $\mathcal{(AB)'}$ |
---|---|---|---|---|---|---|
$0$ | $0$ | $1$ | $1$ | $\LARGE{0}$ | $0$ | $\LARGE{1}$ |
$0$ | $1$ | $1$ | $0$ | $\LARGE{1}$ | $0$ | $\LARGE{1}$ |
$1$ | $0$ | $0$ | $1$ | $\LARGE{1}$ | $0$ | $\LARGE{1}$ |
$1$ | $1$ | $0$ | $0$ | $\LARGE{0}$ | $1$ | $\LARGE{0}$ |
As it is clear from table, both the columns are not same for all possible combinations.
Hence, the expressions are not equivalent.