How to Prove It - Ch2 Sec 2 Exercise 2a
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Question: I am uncertain if my initial formulation of the logical form of the below statement is correct.
How to Prove It (Velleman) Chapter 2, Section 2, Exercise 2a
Negate these statements and then reexpress the results as equivalent positive statements. (it is implied that you put into logical form first based on worked examples in the chapter)
2(a) "There is someone in the freshman class who doesn't have a roommate."
I have written this (prior to negation)
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)]
\ \
F(x): text{x is in the freshman class}\
R(x, y): text{x is roommates with y}\
end{aligned}end{equation}tag{1}$$
I thought this was the correct formulation however, I found two blogs online that formulate the phrase as follows (again, before negation)
$$begin{equation}begin{aligned}
exists x[F(x) wedge negexists yR(x,y)]
end{aligned}end{equation}tag{2}$$
I checked to make sure equations (1) and (2) are not equivalent
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)] qquad &text{(1)}\
exists x[F(x) rightarrow neg exists yR(x,y)] qquad &text{quantifier negation}\
exists x[neg F(x) vee neg exists yR(x,y)] qquad &text{conditional law}\
exists xneg[F(x) wedge exists yR(x,y)] qquad &text{DeMorgans}\
negforall x[F(x) wedge exists yR(x,y)] qquad &text{quantifier negation}\
end{aligned}end{equation}$$
I am trying to convince myself that (2) is correct, but can't see why (1) is incorrect. Is it because it's somewhat speculative ("There exists someone x, where if that person x is a freshman then for all people y, person x and person y are not roommates") vs declarative ("There is someone who is a freshman and for all people y, that person and person y are not roommates")?
elementary-set-theory logic proof-writing quantifiers
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add a comment |
$begingroup$
Question: I am uncertain if my initial formulation of the logical form of the below statement is correct.
How to Prove It (Velleman) Chapter 2, Section 2, Exercise 2a
Negate these statements and then reexpress the results as equivalent positive statements. (it is implied that you put into logical form first based on worked examples in the chapter)
2(a) "There is someone in the freshman class who doesn't have a roommate."
I have written this (prior to negation)
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)]
\ \
F(x): text{x is in the freshman class}\
R(x, y): text{x is roommates with y}\
end{aligned}end{equation}tag{1}$$
I thought this was the correct formulation however, I found two blogs online that formulate the phrase as follows (again, before negation)
$$begin{equation}begin{aligned}
exists x[F(x) wedge negexists yR(x,y)]
end{aligned}end{equation}tag{2}$$
I checked to make sure equations (1) and (2) are not equivalent
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)] qquad &text{(1)}\
exists x[F(x) rightarrow neg exists yR(x,y)] qquad &text{quantifier negation}\
exists x[neg F(x) vee neg exists yR(x,y)] qquad &text{conditional law}\
exists xneg[F(x) wedge exists yR(x,y)] qquad &text{DeMorgans}\
negforall x[F(x) wedge exists yR(x,y)] qquad &text{quantifier negation}\
end{aligned}end{equation}$$
I am trying to convince myself that (2) is correct, but can't see why (1) is incorrect. Is it because it's somewhat speculative ("There exists someone x, where if that person x is a freshman then for all people y, person x and person y are not roommates") vs declarative ("There is someone who is a freshman and for all people y, that person and person y are not roommates")?
elementary-set-theory logic proof-writing quantifiers
$endgroup$
$begingroup$
Since you can push the $neg$ in turning $negexists y$ into $forall yneg$, it's worth noting that the only difference between (1) and (2) is $to$ versus $land$.
$endgroup$
– Derek Elkins
Jan 7 at 2:22
add a comment |
$begingroup$
Question: I am uncertain if my initial formulation of the logical form of the below statement is correct.
How to Prove It (Velleman) Chapter 2, Section 2, Exercise 2a
Negate these statements and then reexpress the results as equivalent positive statements. (it is implied that you put into logical form first based on worked examples in the chapter)
2(a) "There is someone in the freshman class who doesn't have a roommate."
I have written this (prior to negation)
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)]
\ \
F(x): text{x is in the freshman class}\
R(x, y): text{x is roommates with y}\
end{aligned}end{equation}tag{1}$$
I thought this was the correct formulation however, I found two blogs online that formulate the phrase as follows (again, before negation)
$$begin{equation}begin{aligned}
exists x[F(x) wedge negexists yR(x,y)]
end{aligned}end{equation}tag{2}$$
I checked to make sure equations (1) and (2) are not equivalent
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)] qquad &text{(1)}\
exists x[F(x) rightarrow neg exists yR(x,y)] qquad &text{quantifier negation}\
exists x[neg F(x) vee neg exists yR(x,y)] qquad &text{conditional law}\
exists xneg[F(x) wedge exists yR(x,y)] qquad &text{DeMorgans}\
negforall x[F(x) wedge exists yR(x,y)] qquad &text{quantifier negation}\
end{aligned}end{equation}$$
I am trying to convince myself that (2) is correct, but can't see why (1) is incorrect. Is it because it's somewhat speculative ("There exists someone x, where if that person x is a freshman then for all people y, person x and person y are not roommates") vs declarative ("There is someone who is a freshman and for all people y, that person and person y are not roommates")?
elementary-set-theory logic proof-writing quantifiers
$endgroup$
Question: I am uncertain if my initial formulation of the logical form of the below statement is correct.
How to Prove It (Velleman) Chapter 2, Section 2, Exercise 2a
Negate these statements and then reexpress the results as equivalent positive statements. (it is implied that you put into logical form first based on worked examples in the chapter)
2(a) "There is someone in the freshman class who doesn't have a roommate."
I have written this (prior to negation)
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)]
\ \
F(x): text{x is in the freshman class}\
R(x, y): text{x is roommates with y}\
end{aligned}end{equation}tag{1}$$
I thought this was the correct formulation however, I found two blogs online that formulate the phrase as follows (again, before negation)
$$begin{equation}begin{aligned}
exists x[F(x) wedge negexists yR(x,y)]
end{aligned}end{equation}tag{2}$$
I checked to make sure equations (1) and (2) are not equivalent
$$begin{equation}begin{aligned}
exists x[F(x) rightarrow forall y neg R(x,y)] qquad &text{(1)}\
exists x[F(x) rightarrow neg exists yR(x,y)] qquad &text{quantifier negation}\
exists x[neg F(x) vee neg exists yR(x,y)] qquad &text{conditional law}\
exists xneg[F(x) wedge exists yR(x,y)] qquad &text{DeMorgans}\
negforall x[F(x) wedge exists yR(x,y)] qquad &text{quantifier negation}\
end{aligned}end{equation}$$
I am trying to convince myself that (2) is correct, but can't see why (1) is incorrect. Is it because it's somewhat speculative ("There exists someone x, where if that person x is a freshman then for all people y, person x and person y are not roommates") vs declarative ("There is someone who is a freshman and for all people y, that person and person y are not roommates")?
elementary-set-theory logic proof-writing quantifiers
elementary-set-theory logic proof-writing quantifiers
asked Jan 7 at 1:57
Jake KirschJake Kirsch
687
687
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Since you can push the $neg$ in turning $negexists y$ into $forall yneg$, it's worth noting that the only difference between (1) and (2) is $to$ versus $land$.
$endgroup$
– Derek Elkins
Jan 7 at 2:22
add a comment |
$begingroup$
Since you can push the $neg$ in turning $negexists y$ into $forall yneg$, it's worth noting that the only difference between (1) and (2) is $to$ versus $land$.
$endgroup$
– Derek Elkins
Jan 7 at 2:22
$begingroup$
Since you can push the $neg$ in turning $negexists y$ into $forall yneg$, it's worth noting that the only difference between (1) and (2) is $to$ versus $land$.
$endgroup$
– Derek Elkins
Jan 7 at 2:22
$begingroup$
Since you can push the $neg$ in turning $negexists y$ into $forall yneg$, it's worth noting that the only difference between (1) and (2) is $to$ versus $land$.
$endgroup$
– Derek Elkins
Jan 7 at 2:22
add a comment |
3 Answers
3
active
oldest
votes
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You are quite correct in your final paragraph, but perhaps could do with a more concrete argument? Consider a sophomore, say. They're not a freshman. So in fact they satisfy equation (1) - the implication holds true trivially because they are not a freshman at all! That is why (2) is the correct formulation - (1) allows the sophomore. (How many roommates they have doesn't matter at all!)
EDIT: Perhaps I could be clearer. Your equation (1) says 'there exists a student who, if they are a freshman, must not have any roommates'. My point is that a sophomore satisfies this condition, and so this does not correspond to the logical statement you were aiming for.
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$begingroup$
ah I think I'm seeing that now. I got turned down the path of using the conditional based on a worked example that read very similarly. I see now that was incorrect.
$endgroup$
– Jake Kirsch
Jan 7 at 10:31
add a comment |
$begingroup$
This year every freshman has a roommate.
Dan, it so happens, is not a freshman.
If he were a freshman, he wouldn't have a roommate.
Thusly Dan satisfies your statement even though
there is no freshman without a roommate.
$endgroup$
add a comment |
$begingroup$
Universal quantifiers are restricted by a conditional; existential quantifiers are restricted by conjunction". Thus you use form (2), as follows.
"There is someone in the freshman class who doesn't have a roommate."
That is to say: "There exists a person, who, is in the freshman class and does not have a roommate."
$$exists x~(F(x)land lnotexists y~R(x,y))$$
$endgroup$
$begingroup$
It would probably be more pedagogically beneficial to also give some explanation or argument as to why the quantifiers are restricted that way. I suspect the OP ended up with their incorrect formula in part because they applied "rules" that they hadn't thought deeply about.
$endgroup$
– Derek Elkins
Jan 7 at 9:46
$begingroup$
@Graham Kemp - I had not been introduced to the concept of restricted quantifiers and their equivalences. Very helpful.
$endgroup$
– Jake Kirsch
Jan 7 at 10:25
add a comment |
Your Answer
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
You are quite correct in your final paragraph, but perhaps could do with a more concrete argument? Consider a sophomore, say. They're not a freshman. So in fact they satisfy equation (1) - the implication holds true trivially because they are not a freshman at all! That is why (2) is the correct formulation - (1) allows the sophomore. (How many roommates they have doesn't matter at all!)
EDIT: Perhaps I could be clearer. Your equation (1) says 'there exists a student who, if they are a freshman, must not have any roommates'. My point is that a sophomore satisfies this condition, and so this does not correspond to the logical statement you were aiming for.
$endgroup$
$begingroup$
ah I think I'm seeing that now. I got turned down the path of using the conditional based on a worked example that read very similarly. I see now that was incorrect.
$endgroup$
– Jake Kirsch
Jan 7 at 10:31
add a comment |
$begingroup$
You are quite correct in your final paragraph, but perhaps could do with a more concrete argument? Consider a sophomore, say. They're not a freshman. So in fact they satisfy equation (1) - the implication holds true trivially because they are not a freshman at all! That is why (2) is the correct formulation - (1) allows the sophomore. (How many roommates they have doesn't matter at all!)
EDIT: Perhaps I could be clearer. Your equation (1) says 'there exists a student who, if they are a freshman, must not have any roommates'. My point is that a sophomore satisfies this condition, and so this does not correspond to the logical statement you were aiming for.
$endgroup$
$begingroup$
ah I think I'm seeing that now. I got turned down the path of using the conditional based on a worked example that read very similarly. I see now that was incorrect.
$endgroup$
– Jake Kirsch
Jan 7 at 10:31
add a comment |
$begingroup$
You are quite correct in your final paragraph, but perhaps could do with a more concrete argument? Consider a sophomore, say. They're not a freshman. So in fact they satisfy equation (1) - the implication holds true trivially because they are not a freshman at all! That is why (2) is the correct formulation - (1) allows the sophomore. (How many roommates they have doesn't matter at all!)
EDIT: Perhaps I could be clearer. Your equation (1) says 'there exists a student who, if they are a freshman, must not have any roommates'. My point is that a sophomore satisfies this condition, and so this does not correspond to the logical statement you were aiming for.
$endgroup$
You are quite correct in your final paragraph, but perhaps could do with a more concrete argument? Consider a sophomore, say. They're not a freshman. So in fact they satisfy equation (1) - the implication holds true trivially because they are not a freshman at all! That is why (2) is the correct formulation - (1) allows the sophomore. (How many roommates they have doesn't matter at all!)
EDIT: Perhaps I could be clearer. Your equation (1) says 'there exists a student who, if they are a freshman, must not have any roommates'. My point is that a sophomore satisfies this condition, and so this does not correspond to the logical statement you were aiming for.
edited Jan 7 at 2:22
answered Jan 7 at 2:13
bouncebackbounceback
444212
444212
$begingroup$
ah I think I'm seeing that now. I got turned down the path of using the conditional based on a worked example that read very similarly. I see now that was incorrect.
$endgroup$
– Jake Kirsch
Jan 7 at 10:31
add a comment |
$begingroup$
ah I think I'm seeing that now. I got turned down the path of using the conditional based on a worked example that read very similarly. I see now that was incorrect.
$endgroup$
– Jake Kirsch
Jan 7 at 10:31
$begingroup$
ah I think I'm seeing that now. I got turned down the path of using the conditional based on a worked example that read very similarly. I see now that was incorrect.
$endgroup$
– Jake Kirsch
Jan 7 at 10:31
$begingroup$
ah I think I'm seeing that now. I got turned down the path of using the conditional based on a worked example that read very similarly. I see now that was incorrect.
$endgroup$
– Jake Kirsch
Jan 7 at 10:31
add a comment |
$begingroup$
This year every freshman has a roommate.
Dan, it so happens, is not a freshman.
If he were a freshman, he wouldn't have a roommate.
Thusly Dan satisfies your statement even though
there is no freshman without a roommate.
$endgroup$
add a comment |
$begingroup$
This year every freshman has a roommate.
Dan, it so happens, is not a freshman.
If he were a freshman, he wouldn't have a roommate.
Thusly Dan satisfies your statement even though
there is no freshman without a roommate.
$endgroup$
add a comment |
$begingroup$
This year every freshman has a roommate.
Dan, it so happens, is not a freshman.
If he were a freshman, he wouldn't have a roommate.
Thusly Dan satisfies your statement even though
there is no freshman without a roommate.
$endgroup$
This year every freshman has a roommate.
Dan, it so happens, is not a freshman.
If he were a freshman, he wouldn't have a roommate.
Thusly Dan satisfies your statement even though
there is no freshman without a roommate.
answered Jan 7 at 2:28
William ElliotWilliam Elliot
8,8232820
8,8232820
add a comment |
add a comment |
$begingroup$
Universal quantifiers are restricted by a conditional; existential quantifiers are restricted by conjunction". Thus you use form (2), as follows.
"There is someone in the freshman class who doesn't have a roommate."
That is to say: "There exists a person, who, is in the freshman class and does not have a roommate."
$$exists x~(F(x)land lnotexists y~R(x,y))$$
$endgroup$
$begingroup$
It would probably be more pedagogically beneficial to also give some explanation or argument as to why the quantifiers are restricted that way. I suspect the OP ended up with their incorrect formula in part because they applied "rules" that they hadn't thought deeply about.
$endgroup$
– Derek Elkins
Jan 7 at 9:46
$begingroup$
@Graham Kemp - I had not been introduced to the concept of restricted quantifiers and their equivalences. Very helpful.
$endgroup$
– Jake Kirsch
Jan 7 at 10:25
add a comment |
$begingroup$
Universal quantifiers are restricted by a conditional; existential quantifiers are restricted by conjunction". Thus you use form (2), as follows.
"There is someone in the freshman class who doesn't have a roommate."
That is to say: "There exists a person, who, is in the freshman class and does not have a roommate."
$$exists x~(F(x)land lnotexists y~R(x,y))$$
$endgroup$
$begingroup$
It would probably be more pedagogically beneficial to also give some explanation or argument as to why the quantifiers are restricted that way. I suspect the OP ended up with their incorrect formula in part because they applied "rules" that they hadn't thought deeply about.
$endgroup$
– Derek Elkins
Jan 7 at 9:46
$begingroup$
@Graham Kemp - I had not been introduced to the concept of restricted quantifiers and their equivalences. Very helpful.
$endgroup$
– Jake Kirsch
Jan 7 at 10:25
add a comment |
$begingroup$
Universal quantifiers are restricted by a conditional; existential quantifiers are restricted by conjunction". Thus you use form (2), as follows.
"There is someone in the freshman class who doesn't have a roommate."
That is to say: "There exists a person, who, is in the freshman class and does not have a roommate."
$$exists x~(F(x)land lnotexists y~R(x,y))$$
$endgroup$
Universal quantifiers are restricted by a conditional; existential quantifiers are restricted by conjunction". Thus you use form (2), as follows.
"There is someone in the freshman class who doesn't have a roommate."
That is to say: "There exists a person, who, is in the freshman class and does not have a roommate."
$$exists x~(F(x)land lnotexists y~R(x,y))$$
answered Jan 7 at 8:27
Graham KempGraham Kemp
87.3k43579
87.3k43579
$begingroup$
It would probably be more pedagogically beneficial to also give some explanation or argument as to why the quantifiers are restricted that way. I suspect the OP ended up with their incorrect formula in part because they applied "rules" that they hadn't thought deeply about.
$endgroup$
– Derek Elkins
Jan 7 at 9:46
$begingroup$
@Graham Kemp - I had not been introduced to the concept of restricted quantifiers and their equivalences. Very helpful.
$endgroup$
– Jake Kirsch
Jan 7 at 10:25
add a comment |
$begingroup$
It would probably be more pedagogically beneficial to also give some explanation or argument as to why the quantifiers are restricted that way. I suspect the OP ended up with their incorrect formula in part because they applied "rules" that they hadn't thought deeply about.
$endgroup$
– Derek Elkins
Jan 7 at 9:46
$begingroup$
@Graham Kemp - I had not been introduced to the concept of restricted quantifiers and their equivalences. Very helpful.
$endgroup$
– Jake Kirsch
Jan 7 at 10:25
$begingroup$
It would probably be more pedagogically beneficial to also give some explanation or argument as to why the quantifiers are restricted that way. I suspect the OP ended up with their incorrect formula in part because they applied "rules" that they hadn't thought deeply about.
$endgroup$
– Derek Elkins
Jan 7 at 9:46
$begingroup$
It would probably be more pedagogically beneficial to also give some explanation or argument as to why the quantifiers are restricted that way. I suspect the OP ended up with their incorrect formula in part because they applied "rules" that they hadn't thought deeply about.
$endgroup$
– Derek Elkins
Jan 7 at 9:46
$begingroup$
@Graham Kemp - I had not been introduced to the concept of restricted quantifiers and their equivalences. Very helpful.
$endgroup$
– Jake Kirsch
Jan 7 at 10:25
$begingroup$
@Graham Kemp - I had not been introduced to the concept of restricted quantifiers and their equivalences. Very helpful.
$endgroup$
– Jake Kirsch
Jan 7 at 10:25
add a comment |
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$begingroup$
Since you can push the $neg$ in turning $negexists y$ into $forall yneg$, it's worth noting that the only difference between (1) and (2) is $to$ versus $land$.
$endgroup$
– Derek Elkins
Jan 7 at 2:22