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Q3E

Expert-verifiedFound in: Page 818

Book edition
7th

Author(s)
Kenneth H. Rosen

Pages
808 pages

ISBN
9780073383095

**Show that \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + }}0){\bf{ = }}1\).****Translate the equation in part \((a)\) into a propositional equivalence by changing each \(0\) into an \({\bf{F}}\), each \(1\) into a \(T\), each Boolean sum into a disjunction, each Boolean product into a conjunction, each complementation into a negation, and the equals sign into a propositional equivalence sign.**

a) The given \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\) is proved.

b) The translate of the given equation \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\) is \(({\bf{T}} \wedge {\bf{T}}) \vee (\neg ({\bf{F}} \wedge {\bf{T}}) \vee {\bf{F}}) \equiv {\bf{T}}\)

**The complement of an element: \({\bf{\bar 0 = 1}}\) and \({\bf{\bar 1 = }}0\)**

**The Boolean sum + or \(OR\) is 1 if either term is 1.**

**The Boolean product \( \cdot \) or \(AND\) is 1 if both terms are 1.**

Using the Boolean product and sum

\(\begin{array}{c}(1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + }}0){\bf{ = 1 + (\bar 0 + 0)}}\\{\bf{ = 1 + (1 + 0)}}\\{\bf{ = 1 + 1}}\\{\bf{ = 1}}\end{array}\)

Therefore, you get \((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\).

\((1 \cdot 1){\bf{ + }}(\overline {0 \cdot 1} {\bf{ + 0}}){\bf{ = 1}}\)

Replace 0 by F

Replace 1 by T

Replace + by \( \vee \)

Replace \( \cdot \) by \( \wedge \)

Replace - by \(\neg \) (placed in front of the expression that used to be underneath)

Replace = by \( \equiv \)

Therefore, you get \(({\bf{T}} \wedge {\bf{T}}) \vee (\neg ({\bf{F}} \wedge {\bf{T}}) \vee {\bf{F}}) \equiv {\bf{T}}\).

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