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Chapter 15: Problem 12

Find the values of the Boolean variables \(w, x, y, z\) that satisfy the following system of simultaneous (Boolean) equations. $$ x+\bar{x} y=0 \quad \bar{x} y=\bar{x} z \quad \bar{x} y+\bar{x} \bar{z}+z w=\bar{z} w $$

Short Answer

Expert verified
The values of \(w, x, y, z\) that satisfy the system of simultaneous Boolean equations are all equal to 1.

Step by step solution

01

Solving the first equation

The first equation is \(x+\bar{x} y = 0\). In Boolean algebra, the sum of a variable and its negation is always 0. Therefore, \(\bar{x}y = 0\), which implies that either \(\bar{x} = 0\) (i.e., \(x = 1\)), or \(y = 0\). But we cannot say for sure yet. So, let's move to the next equation.
02

Solving the second equation

The second equation is \(\bar{x} y=\bar{x} z\), which implies either \(\bar{x} = 0\) (i.e., \(x = 1\)), or \(y = z\). This suggests that \(x = 1\), or that both \(y\) and \(z\) have the same value. Keeping these possibilities in mind, let's move to the last equation.
03

Solving the third equation

The last equation is \(\bar{x} y+\bar{x}\bar{z}+z w=\bar{z} w\). This equation simplifies to \(z = 0\), \(w = 1\) or \(y = 1\). So we have \(z = 0\), \(y = 1\), \(x = 1\) or \(w = 1\).
04

Opting the final conclusion

To synthesize these results, we see that \(x = 1\) is consistent across all equations, so we take \(x = 1\) as the final value. The value of \(y = 1\) is compatible with the previous equations. For \(z\), we see that it should be equal to \(y\), so \(z = y = 1\). As for \(w\), the last equation suggests \(w = 1\), so we also take this as the final value. Hence, the solution to the system of Boolean equations is \(w = 1\), \(x = 1\), \(y = 1\), and \(z = 1\).

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Most popular questions from this chapter

Define the closed binary operation \(\odot\) (Exclusive Or) on \(F_{n}\) by \(f \oplus g=f \bar{g}+\bar{f} g\), where \(f, g: B^{n} \rightarrow B\). a) Determine \(f \oplus f, f \oplus \bar{f}, f \oplus 1\), and \(f \oplus 0\). b) Prove or disprove each of the following: i) \(f \oplus g=0 \Rightarrow f=g\) ii) \(f \oplus(g \oplus h)=(f \oplus g) \oplus h\) iii) \(f \oplus g=\bar{f} \oplus \bar{g}\) iv) \(f \oplus g h=(f \oplus g)(f \oplus h)\) v) \(f(g \oplus h)=f g \oplus f h\) vi) \((\overline{f \oplus g})=f \oplus g=f \oplus \bar{g}\) vii) \(f \oplus g=f \oplus h \Rightarrow g=h\)

Let \(a, b, c \in \mathscr{B}\), a Boolean algebra. Prove that \(a b+c=a(b+c)\) if and only if \(c \leq a\).

Let \(w, x\), and \(y\) be Boolean variables where the value of \(x\) is 1. For each of the following Boolean expressions, determine, if possible, the value of the expression. If you cannot determine the value of the expression, then find the number of assignments of values for \(w\) and \(y\) that will result in the value 1 for the expression. a) \(x+x y+w\) b) \(x y+w\) c) \(\bar{x} y+x w\) d) \(\bar{x} y+w\)

If \(I B\) is a Boolean algebra, partially ordered by \(\leq\), and \(x, y \in \mathscr{B}\), what is the dual of the statement " \(x \leq y " ?\)

Let \(g: B^{4} \rightarrow B\) be defined by \(g(w, x, y, z)=(w z+x y z)(x+\bar{x} \bar{y} z)\). a) Find the d.n.f. and c.n.f. for \(g\). b) Write \(g\) as a sum of minterms and as a product of maxterms (utilizing binary labels).
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