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Prove that if 3x(P(x) → Q(x) is true, then VxP(x) → 3xQ(x) is true.

Prove that if 3x(P(x) → Q(x) is true, then VxP(x) → 3xQ(x) is true.

Database System Concepts
Database System Concepts
7th Edition
ISBN:9780078022159
Author:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher:Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Chapter1: Introduction
Section: Chapter Questions
Problem 1PE
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**Title:** Logical Implication Proof **Objective:** Prove that if ∃x(P(x) → Q(x)) is true, then ∀xP(x) → ∃xQ(x) is true. **Explanation:** The given problem involves proving the logical implication between two quantified statements: 1. **Existential Quantifier → Implication Statement:** - ∃x(P(x) → Q(x)): There exists at least one x such that if P(x) is true, then Q(x) is also true. 2. **Universal Quantifier → Existential Implication Statement:** - ∀xP(x) → ∃xQ(x): If P(x) is true for all x, then there exists an x such that Q(x) is true. **Proof Strategy:** - Assume ∃x(P(x) → Q(x)) is true, which means there is at least one specific instance (let's call it x₀) where P(x₀) implies Q(x₀). - To show ∀xP(x) → ∃xQ(x), consider: - If ∀xP(x) is true, that means P is true for every instance. - Since we already have an x₀ where P(x₀) → Q(x₀) is true, and P(x₀) is true (because ∀xP(x)), Q(x₀) must also be true. - Therefore, ∃xQ(x) is satisfied by x₀. This logical reasoning shows that the initial assumption leads us to the conclusion that ∀xP(x) → ∃xQ(x) is indeed true whenever ∃x(P(x) → Q(x)) is true.
Transcribed Image Text:**Title:** Logical Implication Proof **Objective:** Prove that if ∃x(P(x) → Q(x)) is true, then ∀xP(x) → ∃xQ(x) is true. **Explanation:** The given problem involves proving the logical implication between two quantified statements: 1. **Existential Quantifier → Implication Statement:** - ∃x(P(x) → Q(x)): There exists at least one x such that if P(x) is true, then Q(x) is also true. 2. **Universal Quantifier → Existential Implication Statement:** - ∀xP(x) → ∃xQ(x): If P(x) is true for all x, then there exists an x such that Q(x) is true. **Proof Strategy:** - Assume ∃x(P(x) → Q(x)) is true, which means there is at least one specific instance (let's call it x₀) where P(x₀) implies Q(x₀). - To show ∀xP(x) → ∃xQ(x), consider: - If ∀xP(x) is true, that means P is true for every instance. - Since we already have an x₀ where P(x₀) → Q(x₀) is true, and P(x₀) is true (because ∀xP(x)), Q(x₀) must also be true. - Therefore, ∃xQ(x) is satisfied by x₀. This logical reasoning shows that the initial assumption leads us to the conclusion that ∀xP(x) → ∃xQ(x) is indeed true whenever ∃x(P(x) → Q(x)) is true.
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