How to prove that in intuitionistic logic the contrapositive law is disallowed?












0














here are my efforts:



I need to show that $(P supset Q) equiv (neg Q supset neg P)$



for this I need to construct a kripke model. I construct the following model:



$K = ({0,1}, leq , models)$



then I define the followings for $A$ and $B$



$0 nvDash A$ $1 vDash A$



$ 1 nvDash B$ $0 vDash B $



then I say that



$ 1 nvDash (neg B to neg A) to Ato B$



Can somebody say what is wrong with my proof?






logic







share|cite|improve this question




















  • 1




    Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
    – spaceisdarkgreen
    Nov 16 at 21:41
















0














here are my efforts:



I need to show that $(P supset Q) equiv (neg Q supset neg P)$



for this I need to construct a kripke model. I construct the following model:



$K = ({0,1}, leq , models)$



then I define the followings for $A$ and $B$



$0 nvDash A$ $1 vDash A$



$ 1 nvDash B$ $0 vDash B $



then I say that



$ 1 nvDash (neg B to neg A) to Ato B$



Can somebody say what is wrong with my proof?






logic







share|cite|improve this question




















  • 1




    Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
    – spaceisdarkgreen
    Nov 16 at 21:41














0












0








0







here are my efforts:



I need to show that $(P supset Q) equiv (neg Q supset neg P)$



for this I need to construct a kripke model. I construct the following model:



$K = ({0,1}, leq , models)$



then I define the followings for $A$ and $B$



$0 nvDash A$ $1 vDash A$



$ 1 nvDash B$ $0 vDash B $



then I say that



$ 1 nvDash (neg B to neg A) to Ato B$



Can somebody say what is wrong with my proof?






logic







share|cite|improve this question















here are my efforts:



I need to show that $(P supset Q) equiv (neg Q supset neg P)$



for this I need to construct a kripke model. I construct the following model:



$K = ({0,1}, leq , models)$



then I define the followings for $A$ and $B$



$0 nvDash A$ $1 vDash A$



$ 1 nvDash B$ $0 vDash B $



then I say that



$ 1 nvDash (neg B to neg A) to Ato B$



Can somebody say what is wrong with my proof?






logic




logic






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share|cite|improve this question













share|cite|improve this question




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edited Nov 16 at 20:21









Mauro ALLEGRANZA

64.1k448111




64.1k448111










asked Nov 16 at 20:18









jadenperesl

11




11








  • 1




    Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
    – spaceisdarkgreen
    Nov 16 at 21:41














  • 1




    Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
    – spaceisdarkgreen
    Nov 16 at 21:41








1




1




Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41




Well, you can’t just say things, you have to show them. Also in the convention I’m used to, the greatest world is classical (so that would have to be true in $1$). If you gave some indication why you think this is the case, it would be easy to see if you’re using a different convention or just wrong.
– spaceisdarkgreen
Nov 16 at 21:41










1 Answer
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Long comment



You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.



We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.



In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.






share|cite|improve this answer





















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    1 Answer
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    Long comment



    You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.



    We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.



    In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.






    share|cite|improve this answer


























      1














      Long comment



      You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.



      We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.



      In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.






      share|cite|improve this answer
























        1












        1








        1






        Long comment



        You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.



        We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.



        In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.






        share|cite|improve this answer












        Long comment



        You are violating the rules of Kripke semantics : if $B$ is true in "world" $0$ [in symbols : $w_0 Vdash B$], it cannot be false in $1$, i.e. we cannot have $w_1 nVdash B$.



        We need a three-nodes model with $w_0Rw_1$ and $w_0Rw_2$.



        In no node we must have both $A$ and $B$ true, in order to have : $w_0 nVdash (A to B)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 at 12:25









        Mauro ALLEGRANZA

        64.1k448111




        64.1k448111






























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