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(20 pts) Use the rules of inference to show that if Vx(A(x) → B(x)), 3x(B(x) v C(x)) and Vx¬(D(x) V¬A(x)) are true, then 3x(C(x) AD(x)) is also true, where the domains of all the quantifiers are the same.

(20 pts) Use the rules of inference to show that if Vx(A(x) → B(x)), 3x(B(x) v C(x)) and Vx¬(D(x) V¬A(x)) are true, then 3x(C(x) AD(x)) is also true, where the domains of all the quantifiers are the same.

Advanced Engineering Mathematics
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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**Question:** (20 pts) Use the rules of inference to show that if ∀x(A(x) → B(x)), ∃x(¬B(x) ∨ C(x)) and ∀x ¬(D(x) ∨ ¬A(x)) are true, then ∃x(C(x) ∧ ¬D(x)) is also true, where the domains of all the quantifiers are the same. --- **Solution Explanation:** To solve this problem, we'll use several rules of inference within the scope of predicate logic to derive the conclusion. Here are the steps broken down: 1. **Given Statements:** - ∀x(A(x) → B(x)) - For all x, if A(x) is true, then B(x) is true. - ∃x(¬B(x) ∨ C(x)) - There exists an x such that ¬B(x) or C(x) is true. - ∀x ¬(D(x) ∨ ¬A(x)) - For all x, it is not the case that D(x) or ¬A(x) is true. Simplified, this means A(x) ∧ ¬D(x) is true for all x. 2. **Objective:** - Show that ∃x(C(x) ∧ ¬D(x)) is true. - There exists an x such that both C(x) and ¬D(x) are true. 3. **Proof Steps:** - From ∀x(A(x) → B(x)), assume A(x) is true, then B(x) must be true for all x. - ∃x(¬B(x) ∨ C(x)) implies that there exists some x for which ¬B(x) or C(x) holds: - If ¬B(x) is true, C(x) would need to be true for this x to satisfy ∃x(¬B(x) ∨ C(x)). - From ∀x ¬(D(x) ∨ ¬A(x)), we derive that A(x) ∧ ¬D(x) is true for all x. - This gives us that ¬D(x) is true wherever A(x) is true. - Combine these: - We find an x such that A(x) is true and
Transcribed Image Text:**Question:** (20 pts) Use the rules of inference to show that if ∀x(A(x) → B(x)), ∃x(¬B(x) ∨ C(x)) and ∀x ¬(D(x) ∨ ¬A(x)) are true, then ∃x(C(x) ∧ ¬D(x)) is also true, where the domains of all the quantifiers are the same. --- **Solution Explanation:** To solve this problem, we'll use several rules of inference within the scope of predicate logic to derive the conclusion. Here are the steps broken down: 1. **Given Statements:** - ∀x(A(x) → B(x)) - For all x, if A(x) is true, then B(x) is true. - ∃x(¬B(x) ∨ C(x)) - There exists an x such that ¬B(x) or C(x) is true. - ∀x ¬(D(x) ∨ ¬A(x)) - For all x, it is not the case that D(x) or ¬A(x) is true. Simplified, this means A(x) ∧ ¬D(x) is true for all x. 2. **Objective:** - Show that ∃x(C(x) ∧ ¬D(x)) is true. - There exists an x such that both C(x) and ¬D(x) are true. 3. **Proof Steps:** - From ∀x(A(x) → B(x)), assume A(x) is true, then B(x) must be true for all x. - ∃x(¬B(x) ∨ C(x)) implies that there exists some x for which ¬B(x) or C(x) holds: - If ¬B(x) is true, C(x) would need to be true for this x to satisfy ∃x(¬B(x) ∨ C(x)). - From ∀x ¬(D(x) ∨ ¬A(x)), we derive that A(x) ∧ ¬D(x) is true for all x. - This gives us that ¬D(x) is true wherever A(x) is true. - Combine these: - We find an x such that A(x) is true and
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