why you should follow it? [duplicate]












-3















This question already has an answer here:




  • In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?

    21 answers




We learned expression of deduce, i.e. =>.



But now I dont have capable reason for I agree I represent True if assumption is False.



Anybady there having to explain reason for its deduce?



Best regards,










share|cite|improve this question















marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 2




    Do you mean to ask why $P implies Q$ is true when $P$ is false?
    – quid
    Nov 24 at 20:44








  • 1




    Yes, of course.
    – quickybrown
    Nov 24 at 20:47






  • 2




    @quickybrown Then you should have written that - what you've written right now isn't clear at all.
    – Noah Schweber
    Nov 24 at 20:51










  • In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
    – quickybrown
    Nov 24 at 21:02












  • @quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
    – spaceisdarkgreen
    Nov 24 at 22:04
















-3















This question already has an answer here:




  • In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?

    21 answers




We learned expression of deduce, i.e. =>.



But now I dont have capable reason for I agree I represent True if assumption is False.



Anybady there having to explain reason for its deduce?



Best regards,










share|cite|improve this question















marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.











  • 2




    Do you mean to ask why $P implies Q$ is true when $P$ is false?
    – quid
    Nov 24 at 20:44








  • 1




    Yes, of course.
    – quickybrown
    Nov 24 at 20:47






  • 2




    @quickybrown Then you should have written that - what you've written right now isn't clear at all.
    – Noah Schweber
    Nov 24 at 20:51










  • In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
    – quickybrown
    Nov 24 at 21:02












  • @quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
    – spaceisdarkgreen
    Nov 24 at 22:04














-3












-3








-3








This question already has an answer here:




  • In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?

    21 answers




We learned expression of deduce, i.e. =>.



But now I dont have capable reason for I agree I represent True if assumption is False.



Anybady there having to explain reason for its deduce?



Best regards,










share|cite|improve this question
















This question already has an answer here:




  • In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?

    21 answers




We learned expression of deduce, i.e. =>.



But now I dont have capable reason for I agree I represent True if assumption is False.



Anybady there having to explain reason for its deduce?



Best regards,





This question already has an answer here:




  • In classical logic, why is $(pRightarrow q)$ True if both $p$ and $q$ are False?

    21 answers








logic






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 24 at 21:22

























asked Nov 24 at 20:35









quickybrown

13




13




marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.






marked as duplicate by Shaun, gammatester, Hans Lundmark, spaceisdarkgreen, Derek Elkins Nov 24 at 22:17


This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.










  • 2




    Do you mean to ask why $P implies Q$ is true when $P$ is false?
    – quid
    Nov 24 at 20:44








  • 1




    Yes, of course.
    – quickybrown
    Nov 24 at 20:47






  • 2




    @quickybrown Then you should have written that - what you've written right now isn't clear at all.
    – Noah Schweber
    Nov 24 at 20:51










  • In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
    – quickybrown
    Nov 24 at 21:02












  • @quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
    – spaceisdarkgreen
    Nov 24 at 22:04














  • 2




    Do you mean to ask why $P implies Q$ is true when $P$ is false?
    – quid
    Nov 24 at 20:44








  • 1




    Yes, of course.
    – quickybrown
    Nov 24 at 20:47






  • 2




    @quickybrown Then you should have written that - what you've written right now isn't clear at all.
    – Noah Schweber
    Nov 24 at 20:51










  • In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
    – quickybrown
    Nov 24 at 21:02












  • @quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
    – spaceisdarkgreen
    Nov 24 at 22:04








2




2




Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid
Nov 24 at 20:44






Do you mean to ask why $P implies Q$ is true when $P$ is false?
– quid
Nov 24 at 20:44






1




1




Yes, of course.
– quickybrown
Nov 24 at 20:47




Yes, of course.
– quickybrown
Nov 24 at 20:47




2




2




@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51




@quickybrown Then you should have written that - what you've written right now isn't clear at all.
– Noah Schweber
Nov 24 at 20:51












In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02






In my impression, we are going to merge $urcorner P vee Q and P =>Q$. Is this a correct?
– quickybrown
Nov 24 at 21:02














@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04




@quickybrown Yes, in classical logic, $pto q$ and $lnot plor q$ mean the same thing.
– spaceisdarkgreen
Nov 24 at 22:04










2 Answers
2






active

oldest

votes


















0














Here is a table that should help



begin{array}{|c|c|c|}
hline
P & Q & PRightarrow Q \ hline
T & T & T \ hline
T & F & F \ hline
F & T & T \ hline
F & F & T \ hline
end{array}



For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.



As an extension to this exercise try this one to solidify your understanding:



begin{array}{|c|c|c|}
hline
P & Q & QRightarrow P \ hline
T & T & \ hline
T & F & \ hline
F & T & \ hline
F & F & \ hline
end{array}






share|cite|improve this answer





























    0














    One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
    As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.






    share|cite|improve this answer





















    • But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
      – quickybrown
      Nov 24 at 21:17




















    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0














    Here is a table that should help



    begin{array}{|c|c|c|}
    hline
    P & Q & PRightarrow Q \ hline
    T & T & T \ hline
    T & F & F \ hline
    F & T & T \ hline
    F & F & T \ hline
    end{array}



    For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.



    As an extension to this exercise try this one to solidify your understanding:



    begin{array}{|c|c|c|}
    hline
    P & Q & QRightarrow P \ hline
    T & T & \ hline
    T & F & \ hline
    F & T & \ hline
    F & F & \ hline
    end{array}






    share|cite|improve this answer


























      0














      Here is a table that should help



      begin{array}{|c|c|c|}
      hline
      P & Q & PRightarrow Q \ hline
      T & T & T \ hline
      T & F & F \ hline
      F & T & T \ hline
      F & F & T \ hline
      end{array}



      For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.



      As an extension to this exercise try this one to solidify your understanding:



      begin{array}{|c|c|c|}
      hline
      P & Q & QRightarrow P \ hline
      T & T & \ hline
      T & F & \ hline
      F & T & \ hline
      F & F & \ hline
      end{array}






      share|cite|improve this answer
























        0












        0








        0






        Here is a table that should help



        begin{array}{|c|c|c|}
        hline
        P & Q & PRightarrow Q \ hline
        T & T & T \ hline
        T & F & F \ hline
        F & T & T \ hline
        F & F & T \ hline
        end{array}



        For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.



        As an extension to this exercise try this one to solidify your understanding:



        begin{array}{|c|c|c|}
        hline
        P & Q & QRightarrow P \ hline
        T & T & \ hline
        T & F & \ hline
        F & T & \ hline
        F & F & \ hline
        end{array}






        share|cite|improve this answer












        Here is a table that should help



        begin{array}{|c|c|c|}
        hline
        P & Q & PRightarrow Q \ hline
        T & T & T \ hline
        T & F & F \ hline
        F & T & T \ hline
        F & F & T \ hline
        end{array}



        For the first row, $P$ and $Q$ are both assumed true thus $P Rightarrow Q$ will be true. For the second row, $P$ is true and $Q$ is false thus $PRightarrow Q$ will say okay well what is the result of the ending statement? false since $Q$ is the result of the ending statement in $PRightarrow Q$. Same logic in reverse applied to the third row. Latly, $P$ and $Q$ are now both false, thus $PRightarrow Q$ will return true since both statements both have the same assumed false value. I hope that is clear.



        As an extension to this exercise try this one to solidify your understanding:



        begin{array}{|c|c|c|}
        hline
        P & Q & QRightarrow P \ hline
        T & T & \ hline
        T & F & \ hline
        F & T & \ hline
        F & F & \ hline
        end{array}







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 24 at 21:01









        Wolfy

        2,28011038




        2,28011038























            0














            One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
            As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.






            share|cite|improve this answer





















            • But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
              – quickybrown
              Nov 24 at 21:17


















            0














            One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
            As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.






            share|cite|improve this answer





















            • But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
              – quickybrown
              Nov 24 at 21:17
















            0












            0








            0






            One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
            As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.






            share|cite|improve this answer












            One rather intuitive understanding is that the equivalence $Leftrightarrow$ can be understood as a conjunction of $Leftarrow$ and $Rightarrow$ thus $FRightarrow X$ has to be true independent of whether $X$ stands for true or false.
            As long as the implication does not get violated by a clear example that falsifies it ($TRightarrow F$) it is considered to be true.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 at 21:09









            Nodt Greenish

            29813




            29813












            • But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
              – quickybrown
              Nov 24 at 21:17




















            • But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
              – quickybrown
              Nov 24 at 21:17


















            But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
            – quickybrown
            Nov 24 at 21:17






            But if you have given intuitve function pretend F is T and T is F, can you distinguish between F and T. I don't understand.
            – quickybrown
            Nov 24 at 21:17





            Popular posts from this blog

            Plaza Victoria

            Puebla de Zaragoza

            Musa