How to use natural deduction to show $neg (P land Q) vdash neg P lor neg Q$?












3












$begingroup$


How to use natural deduction to show $lnot (P land Q) vdash lnot P lor lnot Q$? I think I need to first assume $neg(neg P lor neg Q)$ and then find a contradiction but I cannot see how to do it. Can anyone help?






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







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$endgroup$












  • $begingroup$
    Do you know De Morgan's laws?
    $endgroup$
    – lightxbulb
    Jan 21 at 17:41












  • $begingroup$
    @lightxbulb yes, it is what the sequent shows.
    $endgroup$
    – S0rin
    Jan 21 at 17:47










  • $begingroup$
    What rules do you have? What proof system are you using? There are many different proof systems .....
    $endgroup$
    – Bram28
    Jan 21 at 19:45










  • $begingroup$
    The statement isn't constructively true, so it is using LEM, so you may as well just learn how to do truth tables in whatever logic you are using.
    $endgroup$
    – DanielV
    Jan 21 at 21:34
















3












$begingroup$


How to use natural deduction to show $lnot (P land Q) vdash lnot P lor lnot Q$? I think I need to first assume $neg(neg P lor neg Q)$ and then find a contradiction but I cannot see how to do it. Can anyone help?






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







share|cite|improve this question











$endgroup$












  • $begingroup$
    Do you know De Morgan's laws?
    $endgroup$
    – lightxbulb
    Jan 21 at 17:41












  • $begingroup$
    @lightxbulb yes, it is what the sequent shows.
    $endgroup$
    – S0rin
    Jan 21 at 17:47










  • $begingroup$
    What rules do you have? What proof system are you using? There are many different proof systems .....
    $endgroup$
    – Bram28
    Jan 21 at 19:45










  • $begingroup$
    The statement isn't constructively true, so it is using LEM, so you may as well just learn how to do truth tables in whatever logic you are using.
    $endgroup$
    – DanielV
    Jan 21 at 21:34














3












3








3





$begingroup$


How to use natural deduction to show $lnot (P land Q) vdash lnot P lor lnot Q$? I think I need to first assume $neg(neg P lor neg Q)$ and then find a contradiction but I cannot see how to do it. Can anyone help?






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







share|cite|improve this question











$endgroup$




How to use natural deduction to show $lnot (P land Q) vdash lnot P lor lnot Q$? I think I need to first assume $neg(neg P lor neg Q)$ and then find a contradiction but I cannot see how to do it. Can anyone help?






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




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edited Jan 21 at 19:05









Taroccoesbrocco

5,64271840




5,64271840










asked Jan 21 at 17:33









S0rinS0rin

25119




25119












  • $begingroup$
    Do you know De Morgan's laws?
    $endgroup$
    – lightxbulb
    Jan 21 at 17:41












  • $begingroup$
    @lightxbulb yes, it is what the sequent shows.
    $endgroup$
    – S0rin
    Jan 21 at 17:47










  • $begingroup$
    What rules do you have? What proof system are you using? There are many different proof systems .....
    $endgroup$
    – Bram28
    Jan 21 at 19:45










  • $begingroup$
    The statement isn't constructively true, so it is using LEM, so you may as well just learn how to do truth tables in whatever logic you are using.
    $endgroup$
    – DanielV
    Jan 21 at 21:34


















  • $begingroup$
    Do you know De Morgan's laws?
    $endgroup$
    – lightxbulb
    Jan 21 at 17:41












  • $begingroup$
    @lightxbulb yes, it is what the sequent shows.
    $endgroup$
    – S0rin
    Jan 21 at 17:47










  • $begingroup$
    What rules do you have? What proof system are you using? There are many different proof systems .....
    $endgroup$
    – Bram28
    Jan 21 at 19:45










  • $begingroup$
    The statement isn't constructively true, so it is using LEM, so you may as well just learn how to do truth tables in whatever logic you are using.
    $endgroup$
    – DanielV
    Jan 21 at 21:34
















$begingroup$
Do you know De Morgan's laws?
$endgroup$
– lightxbulb
Jan 21 at 17:41






$begingroup$
Do you know De Morgan's laws?
$endgroup$
– lightxbulb
Jan 21 at 17:41














$begingroup$
@lightxbulb yes, it is what the sequent shows.
$endgroup$
– S0rin
Jan 21 at 17:47




$begingroup$
@lightxbulb yes, it is what the sequent shows.
$endgroup$
– S0rin
Jan 21 at 17:47












$begingroup$
What rules do you have? What proof system are you using? There are many different proof systems .....
$endgroup$
– Bram28
Jan 21 at 19:45




$begingroup$
What rules do you have? What proof system are you using? There are many different proof systems .....
$endgroup$
– Bram28
Jan 21 at 19:45












$begingroup$
The statement isn't constructively true, so it is using LEM, so you may as well just learn how to do truth tables in whatever logic you are using.
$endgroup$
– DanielV
Jan 21 at 21:34




$begingroup$
The statement isn't constructively true, so it is using LEM, so you may as well just learn how to do truth tables in whatever logic you are using.
$endgroup$
– DanielV
Jan 21 at 21:34










2 Answers
2






active

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1












$begingroup$


How to use natural deduction to show $¬(P∧Q)⊢¬P∨¬Q$? I think I need to first assume
$¬(¬P∨¬Q)$ and then find a contradiction but I cannot see how to do it.




To show that assuming $lnot(lnot Plorlnot Q)$ contradicts the premise of $lnot(Pland Q)$, you clearly need to show that that assumption entails $Pland Q$.



You can show $lnot(lnot Plorlnot Q)$ entails $P$ by reduction to absurdity.   Assume $lnot P$, then use disjunction introduction, negation elimination, negation introduction (to discharge the assumption), and double negation elimination.



The rest should be obvious.   Here's the fitch style skeleton.



$$deffitch#1#2{quadbegin{array}{|l} #1\ hline #2end{array}}fitch{lnot(Pland Q)hspace{10ex}textsf{Premise}}{fitch{lnot(lnot Plorlnot Q)hspace{3ex}textsf{Assumption}}{fitch{lnot Phspace{9ex}textsf{Assumption}}{lnot Plorlnot Qhspace{3ex}textsf{Disjunction Introduction}\bothspace{10.5ex}textsf{Negation Elimination}}\lnotlnot Phspace{11ex}textsf{Negation Introduction}\Phspace{14ex}textsf{Double Negation Elimination}\vdots }\ vdots\lnot Plorlnot Q}$$






share|cite|improve this answer









$endgroup$





















    4












    $begingroup$

    The following is a derivation in natural deduction proving that $lnot(P land Q) vdash lnot P lor lnot Q$.
    The notation $[A]^*$ marks an assumption discharged by the inference rule $*$.
    begin{align}
    dfrac{!!!!!!!!!!!!!!!!!!!!!!!dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{dfrac{dfrac{lnot(P land Q) quad dfrac{[P]^+ quad dfrac{dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{[lnot Q]^{**}}{lnot P lor lnot Q}lor_{i_2}}{bot}lnot_e!!!!!!!!!!!!!!!!!!!!!!!}{Q}text{raa}^{**}!!!!!!!!!!!!!!}{P land Q}land_i!!!!!!!!!!!!!!!!!!!!!}{bot}lnot_e}{lnot P}lnot_i^+}{lnot P lor lnot Q}lor_{i_1}!!!!!!!!}{bot}lnot_e!!!!!!!!!!!}{lnot P lor lnot Q}text{raa}^*
    end{align}



    The idea is a reasoning by contradiction: in addition to the hypothesis $lnot (P land Q)$, we suppose $lnot (lnot P lor lnot Q)$ (the negation of what we want to prove) and we show that this leads to a contradiction. More precisely, we show that $P land Q$ follows from $lnot (lnot P lor lnot Q)$, which contradicts our hypothesis $lnot (P land Q)$.
    To show this, we need some additional assumptions ($lnot Q$ and $P$), which we can discharge at the right moment: this complicates a bit the reasoning.



    Note the application of the inference rule raa (reductio ad absurdum), used twice. Actually, $lnot(P land Q) vdash lnot P lor lnot Q$ cannot be proved without this rule (or another inference rule equivalent to it).






    share|cite|improve this answer











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      2 Answers
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      2 Answers
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      1












      $begingroup$


      How to use natural deduction to show $¬(P∧Q)⊢¬P∨¬Q$? I think I need to first assume
      $¬(¬P∨¬Q)$ and then find a contradiction but I cannot see how to do it.




      To show that assuming $lnot(lnot Plorlnot Q)$ contradicts the premise of $lnot(Pland Q)$, you clearly need to show that that assumption entails $Pland Q$.



      You can show $lnot(lnot Plorlnot Q)$ entails $P$ by reduction to absurdity.   Assume $lnot P$, then use disjunction introduction, negation elimination, negation introduction (to discharge the assumption), and double negation elimination.



      The rest should be obvious.   Here's the fitch style skeleton.



      $$deffitch#1#2{quadbegin{array}{|l} #1\ hline #2end{array}}fitch{lnot(Pland Q)hspace{10ex}textsf{Premise}}{fitch{lnot(lnot Plorlnot Q)hspace{3ex}textsf{Assumption}}{fitch{lnot Phspace{9ex}textsf{Assumption}}{lnot Plorlnot Qhspace{3ex}textsf{Disjunction Introduction}\bothspace{10.5ex}textsf{Negation Elimination}}\lnotlnot Phspace{11ex}textsf{Negation Introduction}\Phspace{14ex}textsf{Double Negation Elimination}\vdots }\ vdots\lnot Plorlnot Q}$$






      share|cite|improve this answer









      $endgroup$


















        1












        $begingroup$


        How to use natural deduction to show $¬(P∧Q)⊢¬P∨¬Q$? I think I need to first assume
        $¬(¬P∨¬Q)$ and then find a contradiction but I cannot see how to do it.




        To show that assuming $lnot(lnot Plorlnot Q)$ contradicts the premise of $lnot(Pland Q)$, you clearly need to show that that assumption entails $Pland Q$.



        You can show $lnot(lnot Plorlnot Q)$ entails $P$ by reduction to absurdity.   Assume $lnot P$, then use disjunction introduction, negation elimination, negation introduction (to discharge the assumption), and double negation elimination.



        The rest should be obvious.   Here's the fitch style skeleton.



        $$deffitch#1#2{quadbegin{array}{|l} #1\ hline #2end{array}}fitch{lnot(Pland Q)hspace{10ex}textsf{Premise}}{fitch{lnot(lnot Plorlnot Q)hspace{3ex}textsf{Assumption}}{fitch{lnot Phspace{9ex}textsf{Assumption}}{lnot Plorlnot Qhspace{3ex}textsf{Disjunction Introduction}\bothspace{10.5ex}textsf{Negation Elimination}}\lnotlnot Phspace{11ex}textsf{Negation Introduction}\Phspace{14ex}textsf{Double Negation Elimination}\vdots }\ vdots\lnot Plorlnot Q}$$






        share|cite|improve this answer









        $endgroup$
















          1












          1








          1





          $begingroup$


          How to use natural deduction to show $¬(P∧Q)⊢¬P∨¬Q$? I think I need to first assume
          $¬(¬P∨¬Q)$ and then find a contradiction but I cannot see how to do it.




          To show that assuming $lnot(lnot Plorlnot Q)$ contradicts the premise of $lnot(Pland Q)$, you clearly need to show that that assumption entails $Pland Q$.



          You can show $lnot(lnot Plorlnot Q)$ entails $P$ by reduction to absurdity.   Assume $lnot P$, then use disjunction introduction, negation elimination, negation introduction (to discharge the assumption), and double negation elimination.



          The rest should be obvious.   Here's the fitch style skeleton.



          $$deffitch#1#2{quadbegin{array}{|l} #1\ hline #2end{array}}fitch{lnot(Pland Q)hspace{10ex}textsf{Premise}}{fitch{lnot(lnot Plorlnot Q)hspace{3ex}textsf{Assumption}}{fitch{lnot Phspace{9ex}textsf{Assumption}}{lnot Plorlnot Qhspace{3ex}textsf{Disjunction Introduction}\bothspace{10.5ex}textsf{Negation Elimination}}\lnotlnot Phspace{11ex}textsf{Negation Introduction}\Phspace{14ex}textsf{Double Negation Elimination}\vdots }\ vdots\lnot Plorlnot Q}$$






          share|cite|improve this answer









          $endgroup$




          How to use natural deduction to show $¬(P∧Q)⊢¬P∨¬Q$? I think I need to first assume
          $¬(¬P∨¬Q)$ and then find a contradiction but I cannot see how to do it.




          To show that assuming $lnot(lnot Plorlnot Q)$ contradicts the premise of $lnot(Pland Q)$, you clearly need to show that that assumption entails $Pland Q$.



          You can show $lnot(lnot Plorlnot Q)$ entails $P$ by reduction to absurdity.   Assume $lnot P$, then use disjunction introduction, negation elimination, negation introduction (to discharge the assumption), and double negation elimination.



          The rest should be obvious.   Here's the fitch style skeleton.



          $$deffitch#1#2{quadbegin{array}{|l} #1\ hline #2end{array}}fitch{lnot(Pland Q)hspace{10ex}textsf{Premise}}{fitch{lnot(lnot Plorlnot Q)hspace{3ex}textsf{Assumption}}{fitch{lnot Phspace{9ex}textsf{Assumption}}{lnot Plorlnot Qhspace{3ex}textsf{Disjunction Introduction}\bothspace{10.5ex}textsf{Negation Elimination}}\lnotlnot Phspace{11ex}textsf{Negation Introduction}\Phspace{14ex}textsf{Double Negation Elimination}\vdots }\ vdots\lnot Plorlnot Q}$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Jan 23 at 23:34









          Graham KempGraham Kemp

          86.4k43479




          86.4k43479























              4












              $begingroup$

              The following is a derivation in natural deduction proving that $lnot(P land Q) vdash lnot P lor lnot Q$.
              The notation $[A]^*$ marks an assumption discharged by the inference rule $*$.
              begin{align}
              dfrac{!!!!!!!!!!!!!!!!!!!!!!!dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{dfrac{dfrac{lnot(P land Q) quad dfrac{[P]^+ quad dfrac{dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{[lnot Q]^{**}}{lnot P lor lnot Q}lor_{i_2}}{bot}lnot_e!!!!!!!!!!!!!!!!!!!!!!!}{Q}text{raa}^{**}!!!!!!!!!!!!!!}{P land Q}land_i!!!!!!!!!!!!!!!!!!!!!}{bot}lnot_e}{lnot P}lnot_i^+}{lnot P lor lnot Q}lor_{i_1}!!!!!!!!}{bot}lnot_e!!!!!!!!!!!}{lnot P lor lnot Q}text{raa}^*
              end{align}



              The idea is a reasoning by contradiction: in addition to the hypothesis $lnot (P land Q)$, we suppose $lnot (lnot P lor lnot Q)$ (the negation of what we want to prove) and we show that this leads to a contradiction. More precisely, we show that $P land Q$ follows from $lnot (lnot P lor lnot Q)$, which contradicts our hypothesis $lnot (P land Q)$.
              To show this, we need some additional assumptions ($lnot Q$ and $P$), which we can discharge at the right moment: this complicates a bit the reasoning.



              Note the application of the inference rule raa (reductio ad absurdum), used twice. Actually, $lnot(P land Q) vdash lnot P lor lnot Q$ cannot be proved without this rule (or another inference rule equivalent to it).






              share|cite|improve this answer











              $endgroup$


















                4












                $begingroup$

                The following is a derivation in natural deduction proving that $lnot(P land Q) vdash lnot P lor lnot Q$.
                The notation $[A]^*$ marks an assumption discharged by the inference rule $*$.
                begin{align}
                dfrac{!!!!!!!!!!!!!!!!!!!!!!!dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{dfrac{dfrac{lnot(P land Q) quad dfrac{[P]^+ quad dfrac{dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{[lnot Q]^{**}}{lnot P lor lnot Q}lor_{i_2}}{bot}lnot_e!!!!!!!!!!!!!!!!!!!!!!!}{Q}text{raa}^{**}!!!!!!!!!!!!!!}{P land Q}land_i!!!!!!!!!!!!!!!!!!!!!}{bot}lnot_e}{lnot P}lnot_i^+}{lnot P lor lnot Q}lor_{i_1}!!!!!!!!}{bot}lnot_e!!!!!!!!!!!}{lnot P lor lnot Q}text{raa}^*
                end{align}



                The idea is a reasoning by contradiction: in addition to the hypothesis $lnot (P land Q)$, we suppose $lnot (lnot P lor lnot Q)$ (the negation of what we want to prove) and we show that this leads to a contradiction. More precisely, we show that $P land Q$ follows from $lnot (lnot P lor lnot Q)$, which contradicts our hypothesis $lnot (P land Q)$.
                To show this, we need some additional assumptions ($lnot Q$ and $P$), which we can discharge at the right moment: this complicates a bit the reasoning.



                Note the application of the inference rule raa (reductio ad absurdum), used twice. Actually, $lnot(P land Q) vdash lnot P lor lnot Q$ cannot be proved without this rule (or another inference rule equivalent to it).






                share|cite|improve this answer











                $endgroup$
















                  4












                  4








                  4





                  $begingroup$

                  The following is a derivation in natural deduction proving that $lnot(P land Q) vdash lnot P lor lnot Q$.
                  The notation $[A]^*$ marks an assumption discharged by the inference rule $*$.
                  begin{align}
                  dfrac{!!!!!!!!!!!!!!!!!!!!!!!dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{dfrac{dfrac{lnot(P land Q) quad dfrac{[P]^+ quad dfrac{dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{[lnot Q]^{**}}{lnot P lor lnot Q}lor_{i_2}}{bot}lnot_e!!!!!!!!!!!!!!!!!!!!!!!}{Q}text{raa}^{**}!!!!!!!!!!!!!!}{P land Q}land_i!!!!!!!!!!!!!!!!!!!!!}{bot}lnot_e}{lnot P}lnot_i^+}{lnot P lor lnot Q}lor_{i_1}!!!!!!!!}{bot}lnot_e!!!!!!!!!!!}{lnot P lor lnot Q}text{raa}^*
                  end{align}



                  The idea is a reasoning by contradiction: in addition to the hypothesis $lnot (P land Q)$, we suppose $lnot (lnot P lor lnot Q)$ (the negation of what we want to prove) and we show that this leads to a contradiction. More precisely, we show that $P land Q$ follows from $lnot (lnot P lor lnot Q)$, which contradicts our hypothesis $lnot (P land Q)$.
                  To show this, we need some additional assumptions ($lnot Q$ and $P$), which we can discharge at the right moment: this complicates a bit the reasoning.



                  Note the application of the inference rule raa (reductio ad absurdum), used twice. Actually, $lnot(P land Q) vdash lnot P lor lnot Q$ cannot be proved without this rule (or another inference rule equivalent to it).






                  share|cite|improve this answer











                  $endgroup$



                  The following is a derivation in natural deduction proving that $lnot(P land Q) vdash lnot P lor lnot Q$.
                  The notation $[A]^*$ marks an assumption discharged by the inference rule $*$.
                  begin{align}
                  dfrac{!!!!!!!!!!!!!!!!!!!!!!!dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{dfrac{dfrac{lnot(P land Q) quad dfrac{[P]^+ quad dfrac{dfrac{[lnot(lnot P lor lnot Q)]^* quad dfrac{[lnot Q]^{**}}{lnot P lor lnot Q}lor_{i_2}}{bot}lnot_e!!!!!!!!!!!!!!!!!!!!!!!}{Q}text{raa}^{**}!!!!!!!!!!!!!!}{P land Q}land_i!!!!!!!!!!!!!!!!!!!!!}{bot}lnot_e}{lnot P}lnot_i^+}{lnot P lor lnot Q}lor_{i_1}!!!!!!!!}{bot}lnot_e!!!!!!!!!!!}{lnot P lor lnot Q}text{raa}^*
                  end{align}



                  The idea is a reasoning by contradiction: in addition to the hypothesis $lnot (P land Q)$, we suppose $lnot (lnot P lor lnot Q)$ (the negation of what we want to prove) and we show that this leads to a contradiction. More precisely, we show that $P land Q$ follows from $lnot (lnot P lor lnot Q)$, which contradicts our hypothesis $lnot (P land Q)$.
                  To show this, we need some additional assumptions ($lnot Q$ and $P$), which we can discharge at the right moment: this complicates a bit the reasoning.



                  Note the application of the inference rule raa (reductio ad absurdum), used twice. Actually, $lnot(P land Q) vdash lnot P lor lnot Q$ cannot be proved without this rule (or another inference rule equivalent to it).







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Jan 24 at 13:45

























                  answered Jan 21 at 19:05









                  TaroccoesbroccoTaroccoesbrocco

                  5,64271840




                  5,64271840






























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                      1 I tried to upgrade a yummy sandwich made of two test slices (@JsonTest and @JdbcTest in my case, crunchy test code in between) adding spring boot 2.1 flavour to it. But it seems it was not much of a success. I cannot annotate my tests with many @...Test since they are now each bringing their own XxxTestContextBootstrapper. It used to work when they all used same SpringBootTestContextBootstrapper. @RunWith(SpringRunner.class) @JdbcTest @JsonTest public class Test { @Test public void test() { System.out.printn("Hello, World !"); } } The error I get from BootstrapUtils is illegalStateException : Configuration error: found multiple declarations of @BootstrapWith for test class I understand I might be doing something wrong here but is there an easy way I could load both Json and Jdbc contex...
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