Database System Concepts
Database System Concepts
7th Edition
ISBN: 9780078022159
Author: Abraham Silberschatz Professor, Henry F. Korth, S. Sudarshan
Publisher: McGraw-Hill Education
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Bounded Summation
In mathematics, the summation is the process of adding a sequence of numeric values, known as addends and their result is called sum or total. The term summation refers to the addition or sum of two or more numbers or units. For example-
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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.
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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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