How to Prove It - Ch2 Sec 2 Exercise 2a












2












$begingroup$


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")?










share|cite|improve this question









$endgroup$












  • $begingroup$
    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
















2












$begingroup$


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")?










share|cite|improve this question









$endgroup$












  • $begingroup$
    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














2












2








2


1



$begingroup$


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")?










share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











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share|cite|improve this question










asked Jan 7 at 1:57









Jake KirschJake Kirsch

687




687












  • $begingroup$
    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




$begingroup$
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










3 Answers
3






active

oldest

votes


















0












$begingroup$

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.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    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



















0












$begingroup$

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.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    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."



    $$exists x~(F(x)land lnotexists y~R(x,y))$$






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      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











    Your Answer





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    3 Answers
    3






    active

    oldest

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    3 Answers
    3






    active

    oldest

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    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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
















    0












    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      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














    0












    0








    0





    $begingroup$

    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.






    share|cite|improve this answer











    $endgroup$



    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.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    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


















    • $begingroup$
      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











    0












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      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.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        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.






        share|cite|improve this answer









        $endgroup$



        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.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 2:28









        William ElliotWilliam Elliot

        8,8232820




        8,8232820























            0












            $begingroup$

            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."



            $$exists x~(F(x)land lnotexists y~R(x,y))$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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
















            0












            $begingroup$

            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."



            $$exists x~(F(x)land lnotexists y~R(x,y))$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              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














            0












            0








            0





            $begingroup$

            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."



            $$exists x~(F(x)land lnotexists y~R(x,y))$$






            share|cite|improve this answer









            $endgroup$



            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."



            $$exists x~(F(x)land lnotexists y~R(x,y))$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            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


















            • $begingroup$
              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


















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