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To illustrate how you might approach assignment 2, question 1, here is an example. Statement: Let's show that the following statement does NOT hold. Єx € D, (M(x)) ^ ³x E D (P(x)) implies that 3x E D (M(x) ^ P(x)) Intuitively, this statement is not true because the x in both of the quantifiers in the first statement are not necessarily the same! So just because there is an element such that M is true and an element such that P is true, doesn't mean that there is one element where both of them is true. So, let's construct a set of elements that makes this true. Let D = {a, b}, and let M, P be defined on D: x P(x) M(x) a T F b F T So, we can verify the statements: • 3x E D, (M(x)) - yes, verified by the element b. • • 3x E D, ((P(x)) - yes, verified by the element a. ³x € D, (M(x)) ^ ³x E D (P(x)) - yes, as both of the previous two statements show. ³x € D, (M(x) ^ P(x)) – false. There is no element where both M and P are true. Therefore, the implication does not hold, as requested. Predicate Logic 1. Consider each of the following three statements. You can assume for each that D is a set and P and M are predicates over D. Show that each of the statements are not true. You should show this by giving an example of a situation where they are not equivalent, through providing a specific (small, non-empty) set D of elements and the values (T/F) of the predicates on each element of D. You do not have to define how P and M work, just what the result of the predicates are on each element of D. Be sure to explain your choices of set and predicate. a. (3x € D, P(x))^(3x € D, (P(x) = M(x))) implies that (3x = D, (P(x)^M(x))) b. vx € D, (M(x) = P(x)) implies that (3x = D‚¬M(x))^(3x = D, (M(x) ^ P(x)) c. (x = D. (M(x))) = (x = D, (P(x))) implies that vx = D, (M(x) = P(x)) C. ED

To illustrate how you might approach assignment 2, question 1, here is an example. Statement: Let's show that the following statement does NOT hold. Єx € D, (M(x)) ^ ³x E D (P(x)) implies that 3x E D (M(x) ^ P(x)) Intuitively, this statement is not true because the x in both of the quantifiers in the first statement are not necessarily the same! So just because there is an element such that M is true and an element such that P is true, doesn't mean that there is one element where both of them is true. So, let's construct a set of elements that makes this true. Let D = {a, b}, and let M, P be defined on D: x P(x) M(x) a T F b F T So, we can verify the statements: • 3x E D, (M(x)) - yes, verified by the element b. • • 3x E D, ((P(x)) - yes, verified by the element a. ³x € D, (M(x)) ^ ³x E D (P(x)) - yes, as both of the previous two statements show. ³x € D, (M(x) ^ P(x)) – false. There is no element where both M and P are true. Therefore, the implication does not hold, as requested. Predicate Logic 1. Consider each of the following three statements. You can assume for each that D is a set and P and M are predicates over D. Show that each of the statements are not true. You should show this by giving an example of a situation where they are not equivalent, through providing a specific (small, non-empty) set D of elements and the values (T/F) of the predicates on each element of D. You do not have to define how P and M work, just what the result of the predicates are on each element of D. Be sure to explain your choices of set and predicate. a. (3x € D, P(x))^(3x € D, (P(x) = M(x))) implies that (3x = D, (P(x)^M(x))) b. vx € D, (M(x) = P(x)) implies that (3x = D‚¬M(x))^(3x = D, (M(x) ^ P(x)) c. (x = D. (M(x))) = (x = D, (P(x))) implies that vx = D, (M(x) = P(x)) C. ED

Elementary Geometry For College Students, 7e
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 16CT: Let P represent any statement. Classify as true or false. a P and P b P or P
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This is the question with an example solution for it. Show all your work.

To illustrate how you might approach assignment 2, question 1, here is an example. Statement: Let's show that the following statement does NOT hold. Єx € D, (M(x)) ^ ³x E D (P(x)) implies that 3x E D (M(x) ^ P(x)) Intuitively, this statement is not true because the x in both of the quantifiers in the first statement are not necessarily the same! So just because there is an element such that M is true and an element such that P is true, doesn't mean that there is one element where both of them is true. So, let's construct a set of elements that makes this true. Let D = {a, b}, and let M, P be defined on D: x P(x) M(x) a T F b F T So, we can verify the statements: • 3x E D, (M(x)) - yes, verified by the element b. • • 3x E D, ((P(x)) - yes, verified by the element a. ³x € D, (M(x)) ^ ³x E D (P(x)) - yes, as both of the previous two statements show. ³x € D, (M(x) ^ P(x)) – false. There is no element where both M and P are true. Therefore, the implication does not hold, as requested.
Transcribed Image Text:To illustrate how you might approach assignment 2, question 1, here is an example. Statement: Let's show that the following statement does NOT hold. Єx € D, (M(x)) ^ ³x E D (P(x)) implies that 3x E D (M(x) ^ P(x)) Intuitively, this statement is not true because the x in both of the quantifiers in the first statement are not necessarily the same! So just because there is an element such that M is true and an element such that P is true, doesn't mean that there is one element where both of them is true. So, let's construct a set of elements that makes this true. Let D = {a, b}, and let M, P be defined on D: x P(x) M(x) a T F b F T So, we can verify the statements: • 3x E D, (M(x)) - yes, verified by the element b. • • 3x E D, ((P(x)) - yes, verified by the element a. ³x € D, (M(x)) ^ ³x E D (P(x)) - yes, as both of the previous two statements show. ³x € D, (M(x) ^ P(x)) – false. There is no element where both M and P are true. Therefore, the implication does not hold, as requested.
Predicate Logic 1. Consider each of the following three statements. You can assume for each that D is a set and P and M are predicates over D. Show that each of the statements are not true. You should show this by giving an example of a situation where they are not equivalent, through providing a specific (small, non-empty) set D of elements and the values (T/F) of the predicates on each element of D. You do not have to define how P and M work, just what the result of the predicates are on each element of D. Be sure to explain your choices of set and predicate. a. (3x € D, P(x))^(3x € D, (P(x) = M(x))) implies that (3x = D, (P(x)^M(x))) b. vx € D, (M(x) = P(x)) implies that (3x = D‚¬M(x))^(3x = D, (M(x) ^ P(x)) c. (x = D. (M(x))) = (x = D, (P(x))) implies that vx = D, (M(x) = P(x)) C. ED
Transcribed Image Text:Predicate Logic 1. Consider each of the following three statements. You can assume for each that D is a set and P and M are predicates over D. Show that each of the statements are not true. You should show this by giving an example of a situation where they are not equivalent, through providing a specific (small, non-empty) set D of elements and the values (T/F) of the predicates on each element of D. You do not have to define how P and M work, just what the result of the predicates are on each element of D. Be sure to explain your choices of set and predicate. a. (3x € D, P(x))^(3x € D, (P(x) = M(x))) implies that (3x = D, (P(x)^M(x))) b. vx € D, (M(x) = P(x)) implies that (3x = D‚¬M(x))^(3x = D, (M(x) ^ P(x)) c. (x = D. (M(x))) = (x = D, (P(x))) implies that vx = D, (M(x) = P(x)) C. ED
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