We define a logical connective as follows: P Q is true when both P and Q are false, and it is false otherwise. (We read P↓Q as “P nor Q"). (a) Write a truth table for P Q and check that P Q is logically equivalent to -(PV Q). (b) Check that P+Q=Q↓ P. That is, ↓ is commutative. (e) Show that (PQ) ↓ R and P↓ (QR) are logically inequivalent. That is, is not associative.
We define a logical connective as follows: P Q is true when both P and Q are false, and it is false otherwise. (We read P↓Q as “P nor Q"). (a) Write a truth table for P Q and check that P Q is logically equivalent to -(PV Q). (b) Check that P+Q=Q↓ P. That is, ↓ is commutative. (e) Show that (PQ) ↓ R and P↓ (QR) are logically inequivalent. That is, is not associative.
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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1. We define a logical connective ↓ as follows: \( P \downarrow Q \) is true when both \( P \) and \( Q \) are false, and it is false otherwise. (We read \( P \downarrow Q \) as "P nor Q").
(a) Write a truth table for \( P \downarrow Q \) and check that \( P \downarrow Q \) is logically equivalent to \( \neg(P \vee Q) \).
(b) Check that \( P \downarrow Q \equiv Q \downarrow P \). That is, ↓ is commutative.
(c) Show that \( (P \downarrow Q) \downarrow R \) and \( P \downarrow (Q \downarrow R) \) are logically inequivalent. That is, ↓ is not associative.
(d) Show that the logical connectives ¬, ∧, and ∨ can each be expressed entirely in terms of ↓, without using any other logical connectives. Specifically, prove the following:
i. \( \neg P \equiv (P \downarrow P) \).
ii. \( P \land Q \equiv (P \downarrow P) \downarrow (Q \downarrow Q) \).
iii. \( P \lor Q \equiv (P \downarrow Q) \downarrow (P \downarrow Q) \).
(e) Prove that the logical connective ⇒ can be expressed entirely in terms of ↓. That is, show that the sentence \( P \Rightarrow Q \) is logically equivalent to a sentence involving ↓ and no other logical connectives.
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**Explanations of Points:**
- **Truth Table (Point a)**: The task involves creating a table with columns for \( P \), \( Q \), \( P \downarrow Q \), and \( \neg(P \vee Q) \), showing values for all combinations of \( P \) and \( Q \) being true or false. By comparing, it confirms whether \( P \downarrow Q \) is logically equivalent to \( \neg(P \vee Q) \).
- **Commutativity (Point b)**: The goal is to show that swapping \( P \) and \( Q \) in \( P \down](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F6d729654-bbb5-48b6-93b1-c4482402b3a3%2F84ed25db-be2c-40a3-88b8-39107953222f%2F2kgp23s_processed.jpeg&w=3840&q=75)
Transcribed Image Text:Here is a transcription of the image along with explanations:
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1. We define a logical connective ↓ as follows: \( P \downarrow Q \) is true when both \( P \) and \( Q \) are false, and it is false otherwise. (We read \( P \downarrow Q \) as "P nor Q").
(a) Write a truth table for \( P \downarrow Q \) and check that \( P \downarrow Q \) is logically equivalent to \( \neg(P \vee Q) \).
(b) Check that \( P \downarrow Q \equiv Q \downarrow P \). That is, ↓ is commutative.
(c) Show that \( (P \downarrow Q) \downarrow R \) and \( P \downarrow (Q \downarrow R) \) are logically inequivalent. That is, ↓ is not associative.
(d) Show that the logical connectives ¬, ∧, and ∨ can each be expressed entirely in terms of ↓, without using any other logical connectives. Specifically, prove the following:
i. \( \neg P \equiv (P \downarrow P) \).
ii. \( P \land Q \equiv (P \downarrow P) \downarrow (Q \downarrow Q) \).
iii. \( P \lor Q \equiv (P \downarrow Q) \downarrow (P \downarrow Q) \).
(e) Prove that the logical connective ⇒ can be expressed entirely in terms of ↓. That is, show that the sentence \( P \Rightarrow Q \) is logically equivalent to a sentence involving ↓ and no other logical connectives.
---
**Explanations of Points:**
- **Truth Table (Point a)**: The task involves creating a table with columns for \( P \), \( Q \), \( P \downarrow Q \), and \( \neg(P \vee Q) \), showing values for all combinations of \( P \) and \( Q \) being true or false. By comparing, it confirms whether \( P \downarrow Q \) is logically equivalent to \( \neg(P \vee Q) \).
- **Commutativity (Point b)**: The goal is to show that swapping \( P \) and \( Q \) in \( P \down
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