Determine whether each argument is valid. If the argument is valid, givea proof using the laws of logic. If the argument is invalid, give values for the predicates P and Q over the domain a, b that demonstrate the argument is invalid. a) ∃x(p(x)∧Q(x)) ∴∃x Q(x)∧ ∃P(x) b) ∀x(p(x)∨Q(x)) ∴∀x Q(x)∨ ∀P(x)

Elementary Geometry For College Students, 7e
7th Edition
ISBN:9781337614085
Author:Alexander, Daniel C.; Koeberlein, Geralyn M.
Publisher:Alexander, Daniel C.; Koeberlein, Geralyn M.
ChapterP: Preliminary Concepts
SectionP.CT: Test
Problem 10CT: Statement P and Q are true while R is a false statement. Classify as true or false:...
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Determine whether each argument is valid. If the argument is valid, givea proof using the laws of logic. If the argument is invalid, give values for the predicates P and Q over the domain a, b that demonstrate the argument is invalid.

a)
∃x(p(x)∧Q(x))
∴∃x Q(x)∧ ∃P(x)

b)
∀x(p(x)∨Q(x))
∴∀x Q(x)∨ ∀P(x)

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How would you put the last question answered in a truth table? For example setting particular element variables to the predicates domain? I've attached an image of my textbook explaining how to make the truth table to prove a quantified arguement invalid, but I am struggling to understand the explanation. Could you use the invalid argument of this users original question to explain how and make a truth table? 

Showing an argument with quantified statements is invalid
An argument with quantified statements can be shown to be invalid by defining a domain and predicates for which the hypotheses are all
true but the conclusion is false. For example, consider the invalid argument:
3x P(x)
3x Q(x)
: 3X (P(x) A Q(x))
Suppose that domain is the set {c, d). The two hypotheses, 3x P(x) and 3x Q(x), are both true for the values for P and Q on elements c and d
given in the table. However, the conclusion 3x (P(x) A Q(x)) is false. Therefore, the argument is invalid.
P O
Transcribed Image Text:Showing an argument with quantified statements is invalid An argument with quantified statements can be shown to be invalid by defining a domain and predicates for which the hypotheses are all true but the conclusion is false. For example, consider the invalid argument: 3x P(x) 3x Q(x) : 3X (P(x) A Q(x)) Suppose that domain is the set {c, d). The two hypotheses, 3x P(x) and 3x Q(x), are both true for the values for P and Q on elements c and d given in the table. However, the conclusion 3x (P(x) A Q(x)) is false. Therefore, the argument is invalid. P O
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