Show that the proof rule is not sound and proof question












2












$begingroup$


I'm asked to show that the proof rule



begin{equation}
dfrac{varphi to psi}{lnot varphi to lnot psi}
end{equation}



is not sound.



To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?



I'm also asked to show $vdash p lor lnot p$. I can have $lnot (p lor lnot p) to p land lnot p$ as an assumption. When I try to move from the conclusion upward I get



begin{equation}
dfrac{dfrac{p land ¬p}{p}}{p lor lnot p}
end{equation}
as I try to move toward the assumption, but I don't think that's right because $p lor lnot p$ should conclude $bot$, not $p$. If I try to move from the assumption downward toward the conclusion I'm not sure what to do because for an implication elimination wouldn't I need to have



begin{equation}
lnot(p lor ¬p) to p land lnot p qquadqquad lnot (p lor lnot p)
end{equation}



as an assumption rather than just



begin{equation}
lnot (p lor lnot p) to p land lnot p
end{equation}






logic propositional-calculus proof-theory natural-deduction formal-proofs







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    For your first question, find a pair of values for $phi$ and $psi$ such that the formula above the line is true but the formula below is false. For your second question, see Mauro's answer.
    $endgroup$
    – palmpo
    Jan 6 at 19:37
















2












$begingroup$


I'm asked to show that the proof rule



begin{equation}
dfrac{varphi to psi}{lnot varphi to lnot psi}
end{equation}



is not sound.



To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?



I'm also asked to show $vdash p lor lnot p$. I can have $lnot (p lor lnot p) to p land lnot p$ as an assumption. When I try to move from the conclusion upward I get



begin{equation}
dfrac{dfrac{p land ¬p}{p}}{p lor lnot p}
end{equation}
as I try to move toward the assumption, but I don't think that's right because $p lor lnot p$ should conclude $bot$, not $p$. If I try to move from the assumption downward toward the conclusion I'm not sure what to do because for an implication elimination wouldn't I need to have



begin{equation}
lnot(p lor ¬p) to p land lnot p qquadqquad lnot (p lor lnot p)
end{equation}



as an assumption rather than just



begin{equation}
lnot (p lor lnot p) to p land lnot p
end{equation}






logic propositional-calculus proof-theory natural-deduction formal-proofs







share|cite|improve this question











$endgroup$








  • 3




    $begingroup$
    For your first question, find a pair of values for $phi$ and $psi$ such that the formula above the line is true but the formula below is false. For your second question, see Mauro's answer.
    $endgroup$
    – palmpo
    Jan 6 at 19:37














2












2








2





$begingroup$


I'm asked to show that the proof rule



begin{equation}
dfrac{varphi to psi}{lnot varphi to lnot psi}
end{equation}



is not sound.



To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?



I'm also asked to show $vdash p lor lnot p$. I can have $lnot (p lor lnot p) to p land lnot p$ as an assumption. When I try to move from the conclusion upward I get



begin{equation}
dfrac{dfrac{p land ¬p}{p}}{p lor lnot p}
end{equation}
as I try to move toward the assumption, but I don't think that's right because $p lor lnot p$ should conclude $bot$, not $p$. If I try to move from the assumption downward toward the conclusion I'm not sure what to do because for an implication elimination wouldn't I need to have



begin{equation}
lnot(p lor ¬p) to p land lnot p qquadqquad lnot (p lor lnot p)
end{equation}



as an assumption rather than just



begin{equation}
lnot (p lor lnot p) to p land lnot p
end{equation}






logic propositional-calculus proof-theory natural-deduction formal-proofs







share|cite|improve this question











$endgroup$




I'm asked to show that the proof rule



begin{equation}
dfrac{varphi to psi}{lnot varphi to lnot psi}
end{equation}



is not sound.



To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?



I'm also asked to show $vdash p lor lnot p$. I can have $lnot (p lor lnot p) to p land lnot p$ as an assumption. When I try to move from the conclusion upward I get



begin{equation}
dfrac{dfrac{p land ¬p}{p}}{p lor lnot p}
end{equation}
as I try to move toward the assumption, but I don't think that's right because $p lor lnot p$ should conclude $bot$, not $p$. If I try to move from the assumption downward toward the conclusion I'm not sure what to do because for an implication elimination wouldn't I need to have



begin{equation}
lnot(p lor ¬p) to p land lnot p qquadqquad lnot (p lor lnot p)
end{equation}



as an assumption rather than just



begin{equation}
lnot (p lor lnot p) to p land lnot p
end{equation}






logic propositional-calculus proof-theory natural-deduction formal-proofs




logic propositional-calculus proof-theory natural-deduction formal-proofs






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 6 at 21:25









Taroccoesbrocco

5,64271840




5,64271840










asked Jan 6 at 19:00









Ryan ReynoldsRyan Reynolds

161




161








  • 3




    $begingroup$
    For your first question, find a pair of values for $phi$ and $psi$ such that the formula above the line is true but the formula below is false. For your second question, see Mauro's answer.
    $endgroup$
    – palmpo
    Jan 6 at 19:37














  • 3




    $begingroup$
    For your first question, find a pair of values for $phi$ and $psi$ such that the formula above the line is true but the formula below is false. For your second question, see Mauro's answer.
    $endgroup$
    – palmpo
    Jan 6 at 19:37








3




3




$begingroup$
For your first question, find a pair of values for $phi$ and $psi$ such that the formula above the line is true but the formula below is false. For your second question, see Mauro's answer.
$endgroup$
– palmpo
Jan 6 at 19:37




$begingroup$
For your first question, find a pair of values for $phi$ and $psi$ such that the formula above the line is true but the formula below is false. For your second question, see Mauro's answer.
$endgroup$
– palmpo
Jan 6 at 19:37










1 Answer
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oldest

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$begingroup$

The first question is:




To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?




One would check that the premise implies the conclusion noting any line in the truth table where that is not true. From that line one can build a counter-example.



enter image description here



For the second question, the OP wants to show $⊢p∨¬p$. One is permitted to use this portion of the De Morgan rules: $¬(p∨¬p)→p∧¬p $.



Proceeding the way the OP starts to show this we could derive a proof using a Fitch-style proof checker as follows:



enter image description here






share|cite|improve this answer









$endgroup$













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    0












    $begingroup$

    The first question is:




    To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?




    One would check that the premise implies the conclusion noting any line in the truth table where that is not true. From that line one can build a counter-example.



    enter image description here



    For the second question, the OP wants to show $⊢p∨¬p$. One is permitted to use this portion of the De Morgan rules: $¬(p∨¬p)→p∧¬p $.



    Proceeding the way the OP starts to show this we could derive a proof using a Fitch-style proof checker as follows:



    enter image description here






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      The first question is:




      To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?




      One would check that the premise implies the conclusion noting any line in the truth table where that is not true. From that line one can build a counter-example.



      enter image description here



      For the second question, the OP wants to show $⊢p∨¬p$. One is permitted to use this portion of the De Morgan rules: $¬(p∨¬p)→p∧¬p $.



      Proceeding the way the OP starts to show this we could derive a proof using a Fitch-style proof checker as follows:



      enter image description here






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        The first question is:




        To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?




        One would check that the premise implies the conclusion noting any line in the truth table where that is not true. From that line one can build a counter-example.



        enter image description here



        For the second question, the OP wants to show $⊢p∨¬p$. One is permitted to use this portion of the De Morgan rules: $¬(p∨¬p)→p∧¬p $.



        Proceeding the way the OP starts to show this we could derive a proof using a Fitch-style proof checker as follows:



        enter image description here






        share|cite|improve this answer









        $endgroup$



        The first question is:




        To show this would I just make the truth tables for the statement above the line and below the line and show that they are not equivalent?




        One would check that the premise implies the conclusion noting any line in the truth table where that is not true. From that line one can build a counter-example.



        enter image description here



        For the second question, the OP wants to show $⊢p∨¬p$. One is permitted to use this portion of the De Morgan rules: $¬(p∨¬p)→p∧¬p $.



        Proceeding the way the OP starts to show this we could derive a proof using a Fitch-style proof checker as follows:



        enter image description here







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Feb 26 at 7:00









        Frank HubenyFrank Hubeny

        5042519




        5042519






























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