Discharging an assumption for a theorem












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$begingroup$


So it's my understanding that theorems do not rely on an assumption set. However, this theorem that I have come across seems like I will have to make at least one undischarged assumption.



$$├ (P rightarrow Q) lor (Q lor R)$$



My initial approach was to use arrow introduction, so assume P then eventually discharge it once I had gotten Q. But I just don't see how I can get Q.



The rest of the theorem I can get using wedge introduction once I have one of the sides, I just don't see a way to get them.






proof-verification logic







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  • 1




    $begingroup$
    Are you sure this is correct as written (and not, for example, $(Q implies P) lor (Q lor R)$)? Without further assumptions, this is not always true.
    $endgroup$
    – platty
    Dec 14 '18 at 8:21






  • 1




    $begingroup$
    In general, you are right: a "theorem" of propositional calculus must be a tautology and its proof must end without undischarged assumptions.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:51






  • 1




    $begingroup$
    Specifically, the above formula is not a tautology, and thus you cannot prove it. This means that a valid derivation of it must have some undischarged assumption; foe example : $R$. In fact : $R vdash (P to Q) lor (Q lor R)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:52
















0












$begingroup$


So it's my understanding that theorems do not rely on an assumption set. However, this theorem that I have come across seems like I will have to make at least one undischarged assumption.



$$├ (P rightarrow Q) lor (Q lor R)$$



My initial approach was to use arrow introduction, so assume P then eventually discharge it once I had gotten Q. But I just don't see how I can get Q.



The rest of the theorem I can get using wedge introduction once I have one of the sides, I just don't see a way to get them.






proof-verification logic







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Are you sure this is correct as written (and not, for example, $(Q implies P) lor (Q lor R)$)? Without further assumptions, this is not always true.
    $endgroup$
    – platty
    Dec 14 '18 at 8:21






  • 1




    $begingroup$
    In general, you are right: a "theorem" of propositional calculus must be a tautology and its proof must end without undischarged assumptions.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:51






  • 1




    $begingroup$
    Specifically, the above formula is not a tautology, and thus you cannot prove it. This means that a valid derivation of it must have some undischarged assumption; foe example : $R$. In fact : $R vdash (P to Q) lor (Q lor R)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:52














0












0








0





$begingroup$


So it's my understanding that theorems do not rely on an assumption set. However, this theorem that I have come across seems like I will have to make at least one undischarged assumption.



$$├ (P rightarrow Q) lor (Q lor R)$$



My initial approach was to use arrow introduction, so assume P then eventually discharge it once I had gotten Q. But I just don't see how I can get Q.



The rest of the theorem I can get using wedge introduction once I have one of the sides, I just don't see a way to get them.






proof-verification logic







share|cite|improve this question









$endgroup$




So it's my understanding that theorems do not rely on an assumption set. However, this theorem that I have come across seems like I will have to make at least one undischarged assumption.



$$├ (P rightarrow Q) lor (Q lor R)$$



My initial approach was to use arrow introduction, so assume P then eventually discharge it once I had gotten Q. But I just don't see how I can get Q.



The rest of the theorem I can get using wedge introduction once I have one of the sides, I just don't see a way to get them.






proof-verification logic




proof-verification logic






share|cite|improve this question













share|cite|improve this question











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asked Dec 14 '18 at 8:07









logicboilogicboi

1




1








  • 1




    $begingroup$
    Are you sure this is correct as written (and not, for example, $(Q implies P) lor (Q lor R)$)? Without further assumptions, this is not always true.
    $endgroup$
    – platty
    Dec 14 '18 at 8:21






  • 1




    $begingroup$
    In general, you are right: a "theorem" of propositional calculus must be a tautology and its proof must end without undischarged assumptions.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:51






  • 1




    $begingroup$
    Specifically, the above formula is not a tautology, and thus you cannot prove it. This means that a valid derivation of it must have some undischarged assumption; foe example : $R$. In fact : $R vdash (P to Q) lor (Q lor R)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:52














  • 1




    $begingroup$
    Are you sure this is correct as written (and not, for example, $(Q implies P) lor (Q lor R)$)? Without further assumptions, this is not always true.
    $endgroup$
    – platty
    Dec 14 '18 at 8:21






  • 1




    $begingroup$
    In general, you are right: a "theorem" of propositional calculus must be a tautology and its proof must end without undischarged assumptions.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:51






  • 1




    $begingroup$
    Specifically, the above formula is not a tautology, and thus you cannot prove it. This means that a valid derivation of it must have some undischarged assumption; foe example : $R$. In fact : $R vdash (P to Q) lor (Q lor R)$.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 14 '18 at 8:52








1




1




$begingroup$
Are you sure this is correct as written (and not, for example, $(Q implies P) lor (Q lor R)$)? Without further assumptions, this is not always true.
$endgroup$
– platty
Dec 14 '18 at 8:21




$begingroup$
Are you sure this is correct as written (and not, for example, $(Q implies P) lor (Q lor R)$)? Without further assumptions, this is not always true.
$endgroup$
– platty
Dec 14 '18 at 8:21




1




1




$begingroup$
In general, you are right: a "theorem" of propositional calculus must be a tautology and its proof must end without undischarged assumptions.
$endgroup$
– Mauro ALLEGRANZA
Dec 14 '18 at 8:51




$begingroup$
In general, you are right: a "theorem" of propositional calculus must be a tautology and its proof must end without undischarged assumptions.
$endgroup$
– Mauro ALLEGRANZA
Dec 14 '18 at 8:51




1




1




$begingroup$
Specifically, the above formula is not a tautology, and thus you cannot prove it. This means that a valid derivation of it must have some undischarged assumption; foe example : $R$. In fact : $R vdash (P to Q) lor (Q lor R)$.
$endgroup$
– Mauro ALLEGRANZA
Dec 14 '18 at 8:52




$begingroup$
Specifically, the above formula is not a tautology, and thus you cannot prove it. This means that a valid derivation of it must have some undischarged assumption; foe example : $R$. In fact : $R vdash (P to Q) lor (Q lor R)$.
$endgroup$
– Mauro ALLEGRANZA
Dec 14 '18 at 8:52










1 Answer
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$begingroup$

As noted above, this "theorem" is false: making $P$ true and $Q$ and $R$ false results in $(Prightarrow Q)vee (Qvee R)$ being false.




Separately, it's also worth noting that a theorem with hypotheses can be turned into a theorem without hypotheses:




  • If $varphi_1,...,varphi_nvdashpsi$, then $vdash (varphi_1wedge ...wedgevarphi_n)rightarrowpsi$; this is the deduction theorem.


  • What about if we have $Gammavdash psi$ where $Gamma$ is infinite (so we can't take the conjunction of everything in $Gamma$)? Well, proofs can only involve finitely many propositions, so there is some finite ${varphi_1,...,varphi_n}subseteqGamma$ such that $varphi_1,...,varphi_nvdashpsi$; and now we're in the situation above.



In a previous edit I mentioned the compactness theorem in the context of the second bulletpoint above. This was a slip-up on my part: the compactness theorem is about $models$, not $vdash$. Of course they're actually equivalent, but this question is about $vdash$ instead of $models$ and $vdash$ is "finitary" by definition; so compactness isn't relevant here.






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    $begingroup$

    As noted above, this "theorem" is false: making $P$ true and $Q$ and $R$ false results in $(Prightarrow Q)vee (Qvee R)$ being false.




    Separately, it's also worth noting that a theorem with hypotheses can be turned into a theorem without hypotheses:




    • If $varphi_1,...,varphi_nvdashpsi$, then $vdash (varphi_1wedge ...wedgevarphi_n)rightarrowpsi$; this is the deduction theorem.


    • What about if we have $Gammavdash psi$ where $Gamma$ is infinite (so we can't take the conjunction of everything in $Gamma$)? Well, proofs can only involve finitely many propositions, so there is some finite ${varphi_1,...,varphi_n}subseteqGamma$ such that $varphi_1,...,varphi_nvdashpsi$; and now we're in the situation above.



    In a previous edit I mentioned the compactness theorem in the context of the second bulletpoint above. This was a slip-up on my part: the compactness theorem is about $models$, not $vdash$. Of course they're actually equivalent, but this question is about $vdash$ instead of $models$ and $vdash$ is "finitary" by definition; so compactness isn't relevant here.






    share|cite|improve this answer











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      0












      $begingroup$

      As noted above, this "theorem" is false: making $P$ true and $Q$ and $R$ false results in $(Prightarrow Q)vee (Qvee R)$ being false.




      Separately, it's also worth noting that a theorem with hypotheses can be turned into a theorem without hypotheses:




      • If $varphi_1,...,varphi_nvdashpsi$, then $vdash (varphi_1wedge ...wedgevarphi_n)rightarrowpsi$; this is the deduction theorem.


      • What about if we have $Gammavdash psi$ where $Gamma$ is infinite (so we can't take the conjunction of everything in $Gamma$)? Well, proofs can only involve finitely many propositions, so there is some finite ${varphi_1,...,varphi_n}subseteqGamma$ such that $varphi_1,...,varphi_nvdashpsi$; and now we're in the situation above.



      In a previous edit I mentioned the compactness theorem in the context of the second bulletpoint above. This was a slip-up on my part: the compactness theorem is about $models$, not $vdash$. Of course they're actually equivalent, but this question is about $vdash$ instead of $models$ and $vdash$ is "finitary" by definition; so compactness isn't relevant here.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        As noted above, this "theorem" is false: making $P$ true and $Q$ and $R$ false results in $(Prightarrow Q)vee (Qvee R)$ being false.




        Separately, it's also worth noting that a theorem with hypotheses can be turned into a theorem without hypotheses:




        • If $varphi_1,...,varphi_nvdashpsi$, then $vdash (varphi_1wedge ...wedgevarphi_n)rightarrowpsi$; this is the deduction theorem.


        • What about if we have $Gammavdash psi$ where $Gamma$ is infinite (so we can't take the conjunction of everything in $Gamma$)? Well, proofs can only involve finitely many propositions, so there is some finite ${varphi_1,...,varphi_n}subseteqGamma$ such that $varphi_1,...,varphi_nvdashpsi$; and now we're in the situation above.



        In a previous edit I mentioned the compactness theorem in the context of the second bulletpoint above. This was a slip-up on my part: the compactness theorem is about $models$, not $vdash$. Of course they're actually equivalent, but this question is about $vdash$ instead of $models$ and $vdash$ is "finitary" by definition; so compactness isn't relevant here.






        share|cite|improve this answer











        $endgroup$



        As noted above, this "theorem" is false: making $P$ true and $Q$ and $R$ false results in $(Prightarrow Q)vee (Qvee R)$ being false.




        Separately, it's also worth noting that a theorem with hypotheses can be turned into a theorem without hypotheses:




        • If $varphi_1,...,varphi_nvdashpsi$, then $vdash (varphi_1wedge ...wedgevarphi_n)rightarrowpsi$; this is the deduction theorem.


        • What about if we have $Gammavdash psi$ where $Gamma$ is infinite (so we can't take the conjunction of everything in $Gamma$)? Well, proofs can only involve finitely many propositions, so there is some finite ${varphi_1,...,varphi_n}subseteqGamma$ such that $varphi_1,...,varphi_nvdashpsi$; and now we're in the situation above.



        In a previous edit I mentioned the compactness theorem in the context of the second bulletpoint above. This was a slip-up on my part: the compactness theorem is about $models$, not $vdash$. Of course they're actually equivalent, but this question is about $vdash$ instead of $models$ and $vdash$ is "finitary" by definition; so compactness isn't relevant here.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 14 '18 at 15:25

























        answered Dec 14 '18 at 15:17









        Noah SchweberNoah Schweber

        126k10151290




        126k10151290






























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