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Chapter 6: Problem 5

Simplify the following Boolean functions, stating the law used in each step of the simplification: (a) \(p \cdot(\bar{p}+p \cdot q)\) (b) \(r \cdot(\overline{p+\bar{q} \cdot \bar{r}})\) (c) \((\overline{p \cdot \bar{q}+\bar{p} \cdot q})\) (d) \(p+q+r+\bar{p} \cdot q\) (e) \(\left(\overline{p^{-q}}\right)+(\overline{\bar{p} \cdot q \cdot r})+p\) (f) \(q+p \cdot r+p \cdot q+r\)

Short Answer

Expert verified
(a) \(p \cdot q\), (b) \(r \cdot \bar{p} \cdot q\), (c) \((\bar{p} + q) \cdot (p + \bar{q})\), (d) \(p + q + r\), (e) \(1\), (f) \(q + r\).

Step by step solution

01

Simplify Expression (a)

**Expression (a):** Begin by simplifying the expression \(p \cdot(\bar{p}+p \cdot q)\).- **Apply the Distribution Law:** \(p \cdot \bar{p} + p \cdot p \cdot q\) - **Apply the Complement Law:** \(0 + p \cdot p \cdot q\) - **Apply the Idempotent Law:** \(p \cdot q\) Thus, the simplified expression is \(p \cdot q\).
02

Simplify Expression (b)

**Expression (b):** Simplify \(r \cdot(\overline{p+\bar{q} \cdot \bar{r}})\).- **Apply De Morgan鈥檚 Theorem:** \(r \cdot (\bar{p} \cdot q \cdot r)\) - **Apply the Identity Law:** \(r \cdot \bar{p} \cdot q \cdot r\) - **Apply the Idempotent Law:** \(r \cdot \bar{p} \cdot q\) The simplified expression is \(r \cdot \bar{p} \cdot q\).
03

Simplify Expression (c)

**Expression (c):** Simplify \((\overline{p \cdot \bar{q}+\bar{p} \cdot q})\).- **Apply De Morgan鈥檚 Theorem:** \(\bar{p \cdot \bar{q}} \cdot \bar{\bar{p} \cdot q}\) - **Apply De Morgan鈥檚 Theorem Again:** \((\bar{p} + q) \cdot (p + \bar{q})\) The simplified expression is \((\bar{p} + q) \cdot (p + \bar{q})\).
04

Simplify Expression (d)

**Expression (d):** Simplify \(p+q+r+\bar{p} \cdot q\).- **Apply the Absorption Law:** \(p + q + r\)The simplified expression is \(p + q + r\).
05

Simplify Expression (e)

**Expression (e):** Simplify \((\overline{p^{{-q}}}) + (\overline{\bar{p} \cdot q \cdot r}) + p\) (Note: There seems to be a typo in the expression with \(p^{-q}\) which suggests an operator error but based on logic assume it meant \(p \cdot \bar{q}\)).- **Apply De Morgan's Theorem to Simplify:** \(\bar{p} + q + \bar{p} + \bar{q} + \bar{r} + p\)- **Apply the Consensus Theorem:** \(1\)The simplified expression is \(1\).
06

Simplify Expression (f)

**Expression (f):** Simplify \(q+p \cdot r+p \cdot q+r\).- **Apply the Distribution Law:** \(q+p \cdot (r+q)+r\)- **Use Consensus Theorem:** \(q+p \cdot r+r\)- **Apply Association and Absorption:** \(q + r\)Thus, the simplified expression is \(q + r\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

De Morgan's Theorem
De Morgan's Theorem is a fundamental concept in Boolean Algebra that allows us to simplify expressions involving negations. The theorem provides two transformation rules that can simplify the handling of logical expressions. These rules state:
  • The negation of a disjunction is equivalent to the conjunction of the negations, expressed as:
    \[ \overline{A + B} = \bar{A} \cdot \bar{B} \]
  • The negation of a conjunction is equivalent to the disjunction of the negations, expressed as:
    \[ \overline{A \cdot B} = \bar{A} + \bar{B} \]
Applying De Morgan's Theorem not only aids in switching between AND and OR operations with complements but also simplifies complex expressions, particularly useful in logic circuit design and optimization.
Complement Law
The Complement Law in Boolean Algebra is a straightforward yet powerful tool for simplifying expressions. It states that a variable ANDed with its complement is always equal to zero, and a variable ORed with its complement is always equal to one. These laws are expressed as:
  • AND Operation:
    \[ A \cdot \bar{A} = 0 \]
  • OR Operation:
    \[ A + \bar{A} = 1 \]
This law helps in removing unnecessary parts of expressions when two opposite states rule each other out, and it showcases the principle of mutual exclusivity between a variable and its complement.
Idempotent Law
The Idempotent Law is used in Boolean Algebra to simplify expressions where a variable is repeated. According to this law:
  • AND Operation: A variable ANDed with itself gives the same variable, expressed as:
    \[ A \cdot A = A \]
  • OR Operation: A variable ORed with itself also gives the same variable, expressed as:
    \[ A + A = A \]
These expressions are simplified essentially by recognizing redundant operations. Idempotent Law simplifies expressions by acknowledging that repeating a state does not change its outcome. This comes particularly in handy when simplifying expressions into their most efficient form.
Absorption Law
Absorption Law helps to simplify complex Boolean expressions by eliminating redundant parts of the expression based on dominance in logic. The principle demonstrates:
  • Absorption with AND:
    \[ A + (A \cdot B) = A \]
  • Absorption with OR:
    \[ A \cdot (A + B) = A \]
These laws help to reduce expressions to a simpler form, often revealing the dominant part of the logic which determines the entire expression's behavior. This law eliminates unnecessary terms, helping in deducing essential logical control within systems and expressions.
Distribution Law
The Distribution Law in Boolean Algebra relates to how expressions can be rearranged, mirroring distribution in regular algebra. It defines:
  • AND distributes over OR:
    \[ A \cdot (B + C) = (A \cdot B) + (A \cdot C) \]
  • OR distributes over AND:
    \[ A + (B \cdot C) = (A + B) \cdot (A + C) \]
These rules are pivotal for rearranging and breaking down expressions to simplify them or fit specific logic-circuit configurations. By understanding and utilizing the Distribution Law, expressions in Boolean Algebra can be manipulated to a desired structure that often minimizes hardware requirements in logical implementations.

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