how to prove that (P⊃Q)≡(¬Q⊃¬P) ( P ⊃ Q ) ≡ ( ¬ Q ⊃ ¬ P ) is disallowed in intuitionistic...












2












$begingroup$


this is what I've tried;



define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then



1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )



can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?






logic intuitionistic-logic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:11










  • $begingroup$
    @DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
    $endgroup$
    – katerine
    Dec 5 '18 at 23:33










  • $begingroup$
    I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:37








  • 2




    $begingroup$
    In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 0:57






  • 1




    $begingroup$
    Already asked and answered here.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 6 '18 at 7:45
















2












$begingroup$


this is what I've tried;



define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then



1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )



can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?






logic intuitionistic-logic







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:11










  • $begingroup$
    @DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
    $endgroup$
    – katerine
    Dec 5 '18 at 23:33










  • $begingroup$
    I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:37








  • 2




    $begingroup$
    In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 0:57






  • 1




    $begingroup$
    Already asked and answered here.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 6 '18 at 7:45














2












2








2


1



$begingroup$


this is what I've tried;



define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then



1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )



can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?






logic intuitionistic-logic







share|cite|improve this question











$endgroup$




this is what I've tried;



define kripke model K=({0,1},≤,⊩) where ≤ is the (total) order relation over {0,1} defined by
0≤0 0≤ 11≤1,and ⊩ is a binary relation from {0,1} to the set of propositional variables such that 0⊮A and 1⊩A and 1⊮B and 0⊩B. then



1⊮(A⊃B)≡(¬B⊃¬A) ( B ⊃ A ) ≡ ( ¬ B ⊃ ¬ A )



can somebody tell me if this is correct or if there are errors, what are they and how can they be corrected?






logic intuitionistic-logic




logic intuitionistic-logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 5 '18 at 23:03







katerine

















asked Dec 5 '18 at 21:11









katerinekaterine

405




405








  • 1




    $begingroup$
    A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:11










  • $begingroup$
    @DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
    $endgroup$
    – katerine
    Dec 5 '18 at 23:33










  • $begingroup$
    I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:37








  • 2




    $begingroup$
    In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 0:57






  • 1




    $begingroup$
    Already asked and answered here.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 6 '18 at 7:45














  • 1




    $begingroup$
    A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:11










  • $begingroup$
    @DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
    $endgroup$
    – katerine
    Dec 5 '18 at 23:33










  • $begingroup$
    I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
    $endgroup$
    – Daniel Schepler
    Dec 5 '18 at 23:37








  • 2




    $begingroup$
    In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
    $endgroup$
    – ryan221b
    Dec 6 '18 at 0:57






  • 1




    $begingroup$
    Already asked and answered here.
    $endgroup$
    – Mauro ALLEGRANZA
    Dec 6 '18 at 7:45








1




1




$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11




$begingroup$
A special case of that with $P = top$ would be $Q equiv lnot lnot Q$...
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:11












$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33




$begingroup$
@DanielSchepler P=T? could you please clarify what you mean? I appreciate you trying to help and I need to know how to prove this soon
$endgroup$
– katerine
Dec 5 '18 at 23:33












$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37






$begingroup$
I'm not sure what your question here is. If $P$ is the true proposition, then $(P supset Q) = (top supset Q) equiv Q$ and $(lnot Q supset lnot P) = (lnot Q supset lnot top) equiv (lnot Q supset bot) equiv lnot lnot Q$.
$endgroup$
– Daniel Schepler
Dec 5 '18 at 23:37






2




2




$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57




$begingroup$
In a Kripke model, if $0 Vdash B$ and $0 leq 1$, then $1 Vdash B$ too—theorem’s can’t become “unproved”.
$endgroup$
– ryan221b
Dec 6 '18 at 0:57




1




1




$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45




$begingroup$
Already asked and answered here.
$endgroup$
– Mauro ALLEGRANZA
Dec 6 '18 at 7:45










1 Answer
1






active

oldest

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2












$begingroup$

I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.



Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.



I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:



enter image description here




  • Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.

  • On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.


Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.




You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!






share|cite|improve this answer









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    $begingroup$

    I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.



    Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.



    I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:



    enter image description here




    • Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.

    • On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.


    Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.




    You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.



      Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.



      I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:



      enter image description here




      • Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.

      • On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.


      Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.




      You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.



        Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.



        I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:



        enter image description here




        • Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.

        • On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.


        Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.




        You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!






        share|cite|improve this answer









        $endgroup$



        I am a bit confused by your notation, so I think this may only be a partial answer, but I hope it helps.



        Note that intuitionistically, as classically, $(A supset B) supset (lnot B supset lnot A)$; so to show that the two aren't equivalent, we're going to have to produce a counterexample to $(lnot B supset lnot A) supset (A supset B)$.



        I'm assuming you are familiar with Kripke semantics (quick reminder here: https://math.stackexchange.com/a/3027858/446689). Consider this structure:



        enter image description here




        • Since $1Vdash B$, we have $0notVdashlnot B$ and $1notVdashlnot B$, so $0Vdash (lnot B supset lnot A)$.

        • On the other hand, $0Vdash A$ but $0notVdash B$, so $0notVdash (A supset B)$.


        Therefore, $0notVdash (lnot B supset lnot A) supset (A supset B)$, and so $0notVdash (A supset B) equiv (lnot B supset lnot A)$.




        You also seem to consider $(B supset A) equiv (lnot B supset lnot A)$, but this is even classically false!







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Dec 6 '18 at 2:01









        ryan221bryan221b

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