Contrapositive of $pland q implies r$












1












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Now I understand that strictly speaking the contrapositive of this statement would be $neg r implies neg plorneg q$, but what I would like to do instead is prove that $neg rland q implies neg p$.



What I'm asking is, can I still call this a contrapositive? Or does it have some other name I could use?
Also since I'm doing this for an assignment would it be an acceptable proof of the statement? Or should I take the easy way out and simply assume all three statements and do a contradiction?



The exact wording of the question is "Given that $p$, and that $q$, show that $r$". I can give the actual statements if it helps but I imagine that would be superfluous information.






propositional-calculus







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$endgroup$








  • 1




    $begingroup$
    It's the contrapositive of the "curry'd" version of the statement: $pto (qto r).$
    $endgroup$
    – spaceisdarkgreen
    Jan 12 at 8:08












  • $begingroup$
    lor produces $lor$ and land produces $land$ (and lnot produces $lnot$), as far implications, I always felt that to or rightarrow, both of which produce $to$, were better for propositional formulas compared to implies.
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 8:13










  • $begingroup$
    @drhab: Yes. Back when I was TA'ing basic logic in Beer-Sheva that was my approach: $pto q$ is a proposition, $pimplies q$ is a statement about propositions (and the two are related, of course).
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 9:07
















1












$begingroup$


Now I understand that strictly speaking the contrapositive of this statement would be $neg r implies neg plorneg q$, but what I would like to do instead is prove that $neg rland q implies neg p$.



What I'm asking is, can I still call this a contrapositive? Or does it have some other name I could use?
Also since I'm doing this for an assignment would it be an acceptable proof of the statement? Or should I take the easy way out and simply assume all three statements and do a contradiction?



The exact wording of the question is "Given that $p$, and that $q$, show that $r$". I can give the actual statements if it helps but I imagine that would be superfluous information.






propositional-calculus







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    It's the contrapositive of the "curry'd" version of the statement: $pto (qto r).$
    $endgroup$
    – spaceisdarkgreen
    Jan 12 at 8:08












  • $begingroup$
    lor produces $lor$ and land produces $land$ (and lnot produces $lnot$), as far implications, I always felt that to or rightarrow, both of which produce $to$, were better for propositional formulas compared to implies.
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 8:13










  • $begingroup$
    @drhab: Yes. Back when I was TA'ing basic logic in Beer-Sheva that was my approach: $pto q$ is a proposition, $pimplies q$ is a statement about propositions (and the two are related, of course).
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 9:07














1












1








1





$begingroup$


Now I understand that strictly speaking the contrapositive of this statement would be $neg r implies neg plorneg q$, but what I would like to do instead is prove that $neg rland q implies neg p$.



What I'm asking is, can I still call this a contrapositive? Or does it have some other name I could use?
Also since I'm doing this for an assignment would it be an acceptable proof of the statement? Or should I take the easy way out and simply assume all three statements and do a contradiction?



The exact wording of the question is "Given that $p$, and that $q$, show that $r$". I can give the actual statements if it helps but I imagine that would be superfluous information.






propositional-calculus







share|cite|improve this question











$endgroup$




Now I understand that strictly speaking the contrapositive of this statement would be $neg r implies neg plorneg q$, but what I would like to do instead is prove that $neg rland q implies neg p$.



What I'm asking is, can I still call this a contrapositive? Or does it have some other name I could use?
Also since I'm doing this for an assignment would it be an acceptable proof of the statement? Or should I take the easy way out and simply assume all three statements and do a contradiction?



The exact wording of the question is "Given that $p$, and that $q$, show that $r$". I can give the actual statements if it helps but I imagine that would be superfluous information.






propositional-calculus




propositional-calculus






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 12 at 9:03









J.G.

33.4k23252




33.4k23252










asked Jan 12 at 8:01









S. TeisseireS. Teisseire

112




112








  • 1




    $begingroup$
    It's the contrapositive of the "curry'd" version of the statement: $pto (qto r).$
    $endgroup$
    – spaceisdarkgreen
    Jan 12 at 8:08












  • $begingroup$
    lor produces $lor$ and land produces $land$ (and lnot produces $lnot$), as far implications, I always felt that to or rightarrow, both of which produce $to$, were better for propositional formulas compared to implies.
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 8:13










  • $begingroup$
    @drhab: Yes. Back when I was TA'ing basic logic in Beer-Sheva that was my approach: $pto q$ is a proposition, $pimplies q$ is a statement about propositions (and the two are related, of course).
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 9:07














  • 1




    $begingroup$
    It's the contrapositive of the "curry'd" version of the statement: $pto (qto r).$
    $endgroup$
    – spaceisdarkgreen
    Jan 12 at 8:08












  • $begingroup$
    lor produces $lor$ and land produces $land$ (and lnot produces $lnot$), as far implications, I always felt that to or rightarrow, both of which produce $to$, were better for propositional formulas compared to implies.
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 8:13










  • $begingroup$
    @drhab: Yes. Back when I was TA'ing basic logic in Beer-Sheva that was my approach: $pto q$ is a proposition, $pimplies q$ is a statement about propositions (and the two are related, of course).
    $endgroup$
    – Asaf Karagila♦
    Jan 12 at 9:07








1




1




$begingroup$
It's the contrapositive of the "curry'd" version of the statement: $pto (qto r).$
$endgroup$
– spaceisdarkgreen
Jan 12 at 8:08






$begingroup$
It's the contrapositive of the "curry'd" version of the statement: $pto (qto r).$
$endgroup$
– spaceisdarkgreen
Jan 12 at 8:08














$begingroup$
lor produces $lor$ and land produces $land$ (and lnot produces $lnot$), as far implications, I always felt that to or rightarrow, both of which produce $to$, were better for propositional formulas compared to implies.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:13




$begingroup$
lor produces $lor$ and land produces $land$ (and lnot produces $lnot$), as far implications, I always felt that to or rightarrow, both of which produce $to$, were better for propositional formulas compared to implies.
$endgroup$
– Asaf Karagila♦
Jan 12 at 8:13












$begingroup$
@drhab: Yes. Back when I was TA'ing basic logic in Beer-Sheva that was my approach: $pto q$ is a proposition, $pimplies q$ is a statement about propositions (and the two are related, of course).
$endgroup$
– Asaf Karagila♦
Jan 12 at 9:07




$begingroup$
@drhab: Yes. Back when I was TA'ing basic logic in Beer-Sheva that was my approach: $pto q$ is a proposition, $pimplies q$ is a statement about propositions (and the two are related, of course).
$endgroup$
– Asaf Karagila♦
Jan 12 at 9:07










2 Answers
2






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1












$begingroup$

The contrapositive to $p land qimplies r$ (which is what you are asking for in the title) is $lnot r implies lnot(p land q) = lnot p lor lnot q$. In general, the contrapositive of a statement is that the negation of the consequent implies the negation of the antecedent, however copmlicated the expression (so not only in the case $p implies q$).



This kind of proof is not nessecarily 'worse' in any sense; take this proof from math.stackexchange:



Proposition: $x^4 - x^3 + x^2 neq 1$, then $x neq 1.$



The easy and elegant way to prove this is by contrapositive:



If $x = 1$, then $x^4 - x^3 + x^2 = 1$.



If you only think about what happens when $x^4 - x^3 + x^2 neq 1$, it's not as trivial that $x neq 1.$
There are many cases where contrapositive is much easier than straightforwardly proving the implication; you will become more familiar with when this is so as you do and read more proofs. Also see https://math.stackexchange.com/a/650487/606584.






share|cite|improve this answer









$endgroup$





















    1












    $begingroup$

    Hint



    since $$lnot rLongrightarrow lnot plorlnot q$$then $$lnot rland qLongrightarrow (lnot plorlnot q)land q$$I leave to you to show that $$(lnot plorlnot q)land qequivlnot p$$






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      Since sim is meant as a relation, it has spacing like that of a relation. Using either {sim} to produce the right spacing or lnotto produce $lnot$ will make this answer much more readable.
      $endgroup$
      – Asaf Karagila♦
      Jan 12 at 9:08










    • $begingroup$
      Thank you for the feedback...
      $endgroup$
      – Mostafa Ayaz
      Jan 12 at 9:09












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    2 Answers
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    1












    $begingroup$

    The contrapositive to $p land qimplies r$ (which is what you are asking for in the title) is $lnot r implies lnot(p land q) = lnot p lor lnot q$. In general, the contrapositive of a statement is that the negation of the consequent implies the negation of the antecedent, however copmlicated the expression (so not only in the case $p implies q$).



    This kind of proof is not nessecarily 'worse' in any sense; take this proof from math.stackexchange:



    Proposition: $x^4 - x^3 + x^2 neq 1$, then $x neq 1.$



    The easy and elegant way to prove this is by contrapositive:



    If $x = 1$, then $x^4 - x^3 + x^2 = 1$.



    If you only think about what happens when $x^4 - x^3 + x^2 neq 1$, it's not as trivial that $x neq 1.$
    There are many cases where contrapositive is much easier than straightforwardly proving the implication; you will become more familiar with when this is so as you do and read more proofs. Also see https://math.stackexchange.com/a/650487/606584.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      The contrapositive to $p land qimplies r$ (which is what you are asking for in the title) is $lnot r implies lnot(p land q) = lnot p lor lnot q$. In general, the contrapositive of a statement is that the negation of the consequent implies the negation of the antecedent, however copmlicated the expression (so not only in the case $p implies q$).



      This kind of proof is not nessecarily 'worse' in any sense; take this proof from math.stackexchange:



      Proposition: $x^4 - x^3 + x^2 neq 1$, then $x neq 1.$



      The easy and elegant way to prove this is by contrapositive:



      If $x = 1$, then $x^4 - x^3 + x^2 = 1$.



      If you only think about what happens when $x^4 - x^3 + x^2 neq 1$, it's not as trivial that $x neq 1.$
      There are many cases where contrapositive is much easier than straightforwardly proving the implication; you will become more familiar with when this is so as you do and read more proofs. Also see https://math.stackexchange.com/a/650487/606584.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        The contrapositive to $p land qimplies r$ (which is what you are asking for in the title) is $lnot r implies lnot(p land q) = lnot p lor lnot q$. In general, the contrapositive of a statement is that the negation of the consequent implies the negation of the antecedent, however copmlicated the expression (so not only in the case $p implies q$).



        This kind of proof is not nessecarily 'worse' in any sense; take this proof from math.stackexchange:



        Proposition: $x^4 - x^3 + x^2 neq 1$, then $x neq 1.$



        The easy and elegant way to prove this is by contrapositive:



        If $x = 1$, then $x^4 - x^3 + x^2 = 1$.



        If you only think about what happens when $x^4 - x^3 + x^2 neq 1$, it's not as trivial that $x neq 1.$
        There are many cases where contrapositive is much easier than straightforwardly proving the implication; you will become more familiar with when this is so as you do and read more proofs. Also see https://math.stackexchange.com/a/650487/606584.






        share|cite|improve this answer









        $endgroup$



        The contrapositive to $p land qimplies r$ (which is what you are asking for in the title) is $lnot r implies lnot(p land q) = lnot p lor lnot q$. In general, the contrapositive of a statement is that the negation of the consequent implies the negation of the antecedent, however copmlicated the expression (so not only in the case $p implies q$).



        This kind of proof is not nessecarily 'worse' in any sense; take this proof from math.stackexchange:



        Proposition: $x^4 - x^3 + x^2 neq 1$, then $x neq 1.$



        The easy and elegant way to prove this is by contrapositive:



        If $x = 1$, then $x^4 - x^3 + x^2 = 1$.



        If you only think about what happens when $x^4 - x^3 + x^2 neq 1$, it's not as trivial that $x neq 1.$
        There are many cases where contrapositive is much easier than straightforwardly proving the implication; you will become more familiar with when this is so as you do and read more proofs. Also see https://math.stackexchange.com/a/650487/606584.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 8:20









        Steven WagterSteven Wagter

        1789




        1789























            1












            $begingroup$

            Hint



            since $$lnot rLongrightarrow lnot plorlnot q$$then $$lnot rland qLongrightarrow (lnot plorlnot q)land q$$I leave to you to show that $$(lnot plorlnot q)land qequivlnot p$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Since sim is meant as a relation, it has spacing like that of a relation. Using either {sim} to produce the right spacing or lnotto produce $lnot$ will make this answer much more readable.
              $endgroup$
              – Asaf Karagila♦
              Jan 12 at 9:08










            • $begingroup$
              Thank you for the feedback...
              $endgroup$
              – Mostafa Ayaz
              Jan 12 at 9:09
















            1












            $begingroup$

            Hint



            since $$lnot rLongrightarrow lnot plorlnot q$$then $$lnot rland qLongrightarrow (lnot plorlnot q)land q$$I leave to you to show that $$(lnot plorlnot q)land qequivlnot p$$






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Since sim is meant as a relation, it has spacing like that of a relation. Using either {sim} to produce the right spacing or lnotto produce $lnot$ will make this answer much more readable.
              $endgroup$
              – Asaf Karagila♦
              Jan 12 at 9:08










            • $begingroup$
              Thank you for the feedback...
              $endgroup$
              – Mostafa Ayaz
              Jan 12 at 9:09














            1












            1








            1





            $begingroup$

            Hint



            since $$lnot rLongrightarrow lnot plorlnot q$$then $$lnot rland qLongrightarrow (lnot plorlnot q)land q$$I leave to you to show that $$(lnot plorlnot q)land qequivlnot p$$






            share|cite|improve this answer











            $endgroup$



            Hint



            since $$lnot rLongrightarrow lnot plorlnot q$$then $$lnot rland qLongrightarrow (lnot plorlnot q)land q$$I leave to you to show that $$(lnot plorlnot q)land qequivlnot p$$







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Jan 12 at 9:09

























            answered Jan 12 at 8:57









            Mostafa AyazMostafa Ayaz

            18.1k31040




            18.1k31040












            • $begingroup$
              Since sim is meant as a relation, it has spacing like that of a relation. Using either {sim} to produce the right spacing or lnotto produce $lnot$ will make this answer much more readable.
              $endgroup$
              – Asaf Karagila♦
              Jan 12 at 9:08










            • $begingroup$
              Thank you for the feedback...
              $endgroup$
              – Mostafa Ayaz
              Jan 12 at 9:09


















            • $begingroup$
              Since sim is meant as a relation, it has spacing like that of a relation. Using either {sim} to produce the right spacing or lnotto produce $lnot$ will make this answer much more readable.
              $endgroup$
              – Asaf Karagila♦
              Jan 12 at 9:08










            • $begingroup$
              Thank you for the feedback...
              $endgroup$
              – Mostafa Ayaz
              Jan 12 at 9:09
















            $begingroup$
            Since sim is meant as a relation, it has spacing like that of a relation. Using either {sim} to produce the right spacing or lnotto produce $lnot$ will make this answer much more readable.
            $endgroup$
            – Asaf Karagila♦
            Jan 12 at 9:08




            $begingroup$
            Since sim is meant as a relation, it has spacing like that of a relation. Using either {sim} to produce the right spacing or lnotto produce $lnot$ will make this answer much more readable.
            $endgroup$
            – Asaf Karagila♦
            Jan 12 at 9:08












            $begingroup$
            Thank you for the feedback...
            $endgroup$
            – Mostafa Ayaz
            Jan 12 at 9:09




            $begingroup$
            Thank you for the feedback...
            $endgroup$
            – Mostafa Ayaz
            Jan 12 at 9:09


















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