i (ΑΠAC) υ (An (AU B))G. ii ¬((¬PV¬Q)^Q)

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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Simplify each of the following expressions. Label each identity used. Be mindful that you use the appropriate notation for set algebra or boolean algebra.

**Logical and Set-Theoretic Laws**

### Associative Laws
- Logical:
  - \((P \lor Q) \lor R = P \lor (Q \lor R)\)
  - \((P \land Q) \land R = P \land (Q \land R)\)

- Set-Theoretic:
  - \((A \cup B) \cup C = A \cup (B \cup C)\)
  - \((A \cap B) \cap C = A \cap (B \cap C)\)

### Double Negation
- Logical: 
  - \(\lnot \lnot P = P\)

- Set-Theoretic: 
  - \((A^C)^C = A\)

### DeMorgan’s Laws
- Logical:
  - \(\lnot (P \lor Q) = \lnot P \land \lnot Q\)
  - \(\lnot (P \land Q) = \lnot P \lor \lnot Q\)

- Set-Theoretic:
  - \((A \cup B)^C = A^C \cap B^C\)
  - \((A \cap B)^C = A^C \cup B^C\)

### Distributive Laws
- Logical:
  - \(P \land (Q \lor R) = (P \land Q) \lor (P \land R)\)
  - \(P \lor (Q \land R) = (P \lor Q) \land (P \lor R)\)

- Set-Theoretic:
  - \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\)
  - \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\)

### Absorption Laws
- Logical:
  - \(P \lor (P \land Q) = P\)
  - \(P \land (P \lor Q) = P\)

- Set-Theoretic:
  - \(A \cap (A \cup B) = A\)
  - \(A \cup (A \cap B) = A\)

### Complement Laws
- Logical:
  - \(P \lor \lnot P = T\)
  - \(P \land \lnot
Transcribed Image Text:**Logical and Set-Theoretic Laws** ### Associative Laws - Logical: - \((P \lor Q) \lor R = P \lor (Q \lor R)\) - \((P \land Q) \land R = P \land (Q \land R)\) - Set-Theoretic: - \((A \cup B) \cup C = A \cup (B \cup C)\) - \((A \cap B) \cap C = A \cap (B \cap C)\) ### Double Negation - Logical: - \(\lnot \lnot P = P\) - Set-Theoretic: - \((A^C)^C = A\) ### DeMorgan’s Laws - Logical: - \(\lnot (P \lor Q) = \lnot P \land \lnot Q\) - \(\lnot (P \land Q) = \lnot P \lor \lnot Q\) - Set-Theoretic: - \((A \cup B)^C = A^C \cap B^C\) - \((A \cap B)^C = A^C \cup B^C\) ### Distributive Laws - Logical: - \(P \land (Q \lor R) = (P \land Q) \lor (P \land R)\) - \(P \lor (Q \land R) = (P \lor Q) \land (P \lor R)\) - Set-Theoretic: - \(A \cap (B \cup C) = (A \cap B) \cup (A \cap C)\) - \(A \cup (B \cap C) = (A \cup B) \cap (A \cup C)\) ### Absorption Laws - Logical: - \(P \lor (P \land Q) = P\) - \(P \land (P \lor Q) = P\) - Set-Theoretic: - \(A \cap (A \cup B) = A\) - \(A \cup (A \cap B) = A\) ### Complement Laws - Logical: - \(P \lor \lnot P = T\) - \(P \land \lnot
The image contains two logical expressions:

i. \[ \left((A \cap A^C) \cup (A \cap (A \cup B))\right)^C \]

ii. \[ \neg\left((\neg P \lor \neg Q) \land Q\right) \]

No graphs or diagrams are present. These expressions involve set operations and logical operators. 

- The first expression utilizes set intersection (\(\cap\)), union (\(\cup\)), and complements (\(^C\)).
- The second expression involves negation (\(\neg\)), logical OR (\(\lor\)), and logical AND (\(\land\)).
Transcribed Image Text:The image contains two logical expressions: i. \[ \left((A \cap A^C) \cup (A \cap (A \cup B))\right)^C \] ii. \[ \neg\left((\neg P \lor \neg Q) \land Q\right) \] No graphs or diagrams are present. These expressions involve set operations and logical operators. - The first expression utilizes set intersection (\(\cap\)), union (\(\cup\)), and complements (\(^C\)). - The second expression involves negation (\(\neg\)), logical OR (\(\lor\)), and logical AND (\(\land\)).
Expert Solution
Step 1

Part i.

AACAABCUse complement law and absorption law:ACUse identity law:ACAC

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