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Expression 1: $Y = B'C' + B'A' + A'C$.

Expression 2: $Y = B'C' + A'C$.

Both of the expressions generate the same truth table which implies they are the same.

However, how do you demonstrate they are the same using just boolean theorem?

If both expressions are the same, then it implies that $B'A' = 0$ and how do you get this?

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Sebastiano
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Anastazia Nichole
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2 Answers2

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It doesn't imply $B'A'=0$. It implies $B'A' \Rightarrow B'C'+A'C$.

If $B'$ and $A'$ are both true, then either $C$ is true, in which case $A'C$ is true, or $C'$ is true, in which case $B'C'$ is true. Thus, $B'A' \Rightarrow B'C'+A'C$.

Robert Shore
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This statement: "... it implies that $B′A′=0$ .." is not correct. It rather means that in case $B′A′$ is $1$ then will at least one of $B′C′$ and $A′C$ also be $1$. Consequently, $B′A′$ won't have any impact on the final result.

It can be proven like this.

Use that ($X'$ + $X$) is always $1$ and that multiplying a term by $1$ doesn't change the result.

Therefor you can rewrite Expression 1 like:

$Y = B'C' + B'A' + A'C$

$Y = (A' + A)B'C' + (C' + C)B'A' + (B' + B)A'C$

$Y = A'B'C' + AB'C' + C'B'A' + CB'A' + B'A'C + BA'C$

Write all terms to be A folowed by B followed by C:

$Y = A'B'C' + AB'C' + A'B'C' + A'B'C + A'B'C + A'BC$

Remove the terms that appear twice:

$Y = A'B'C' + AB'C' + A'B'C + A'BC$

$Y = (A' + A)B'C' + (B' + B)A'C$

$Y = B'C' + A'C$

which is Expression 2.

Another way to look at it is:

Case 1: $B'A'$ is $0$

This is trivial as Expression 1 becomes identical to Expression 2 when $B'A'$ is $0$

Case 2: $B'A'$ is $1$

In this case Expression 1 will give the result $1$

It is also known that $A'$ is $1$ and $B'$ is $1$. This can be inserted into Expression 2 like:

$Y = B′C′ + A′C = 1C' + 1C = C' + C = 1$

So Expression 2 will also give the result $1$

Consequently, Expression 1 and Expression2 will give the same result in all cases.

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