$Q(n)$: “$P(k)$ holds for all $k<n$”. Then why is $Q(0)$ clearly true? [duplicate]












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This question already has an answer here:




  • Strong induction and vacuous truth

    2 answers



  • Strong Induction Requires No Base Case?

    2 answers



  • Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?

    6 answers




It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.



In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.



$Q(0)$ means that "$P(k)$ holds for all $k<0$".



I understood there are no $k<0$.



And then I couldn't proceed.



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marked as duplicate by Asaf Karagila set-theory
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Jan 4 at 12:38


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Vacuous truth
    $endgroup$
    – Wojowu
    Jan 4 at 12:05
















0












$begingroup$



This question already has an answer here:




  • Strong induction and vacuous truth

    2 answers



  • Strong Induction Requires No Base Case?

    2 answers



  • Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?

    6 answers




It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.



In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.



$Q(0)$ means that "$P(k)$ holds for all $k<0$".



I understood there are no $k<0$.



And then I couldn't proceed.



figure for question










share|cite|improve this question











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marked as duplicate by Asaf Karagila set-theory
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Jan 4 at 12:38


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.


















  • $begingroup$
    Vacuous truth
    $endgroup$
    – Wojowu
    Jan 4 at 12:05














0












0








0





$begingroup$



This question already has an answer here:




  • Strong induction and vacuous truth

    2 answers



  • Strong Induction Requires No Base Case?

    2 answers



  • Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?

    6 answers




It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.



In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.



$Q(0)$ means that "$P(k)$ holds for all $k<0$".



I understood there are no $k<0$.



And then I couldn't proceed.



figure for question










share|cite|improve this question











$endgroup$





This question already has an answer here:




  • Strong induction and vacuous truth

    2 answers



  • Strong Induction Requires No Base Case?

    2 answers



  • Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?

    6 answers




It is a principle and proof from Introduction to Set Theory, Hrbacek and Jech.



In the proof, line 1 and 2, I couldn't understand why $Q(0)$ is true.



$Q(0)$ means that "$P(k)$ holds for all $k<0$".



I understood there are no $k<0$.



And then I couldn't proceed.



figure for question





This question already has an answer here:




  • Strong induction and vacuous truth

    2 answers



  • Strong Induction Requires No Base Case?

    2 answers



  • Why is complete strong induction a valid proof method and not need to explicitly proof the base cases?

    6 answers








induction proof-explanation






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edited Jan 4 at 17:54









Andrés E. Caicedo

65.3k8158247




65.3k8158247










asked Jan 4 at 12:02









Doyun NamDoyun Nam

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66119




marked as duplicate by Asaf Karagila set-theory
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Jan 4 at 12:38


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.









marked as duplicate by Asaf Karagila set-theory
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Jan 4 at 12:38


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.














  • $begingroup$
    Vacuous truth
    $endgroup$
    – Wojowu
    Jan 4 at 12:05


















  • $begingroup$
    Vacuous truth
    $endgroup$
    – Wojowu
    Jan 4 at 12:05
















$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05




$begingroup$
Vacuous truth
$endgroup$
– Wojowu
Jan 4 at 12:05










2 Answers
2






active

oldest

votes


















1












$begingroup$

$Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.






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$endgroup$





















    1












    $begingroup$

    A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.



    If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.






    share|cite|improve this answer









    $endgroup$




















      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1












      $begingroup$

      $Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$

        $Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$

          $Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.






          share|cite|improve this answer









          $endgroup$



          $Q(n)$ is the predicate "$forall kin{Bbb N}_0 (k<nRightarrow P(k))$." Then $Q(0)$ is vacuously true, since the premise is false and so the implication is true.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 4 at 12:23









          WuestenfuxWuestenfux

          4,2871413




          4,2871413























              1












              $begingroup$

              A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.



              If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.






              share|cite|improve this answer









              $endgroup$


















                1












                $begingroup$

                A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.



                If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.






                share|cite|improve this answer









                $endgroup$
















                  1












                  1








                  1





                  $begingroup$

                  A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.



                  If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.






                  share|cite|improve this answer









                  $endgroup$



                  A sentence like "for all $x$, if $x$ has the property $A$, then $x$ has the property $B$" is false if (and only if) there exists a counterexample, that is, if there exists some $x$ with the property $A$ but without the property $B$.



                  If there is no $x$ that has the property $A$, then there is no counterexample, so the sentence is true.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Jan 4 at 12:09









                  ajotatxeajotatxe

                  53.8k23890




                  53.8k23890















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