What is a natural deduction proof from ~(A↔B) to ~(A→B)?help with deductive proofUsing natural deduction...

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What is a natural deduction proof from ~(A↔B) to ~(A→B)?


help with deductive proofUsing natural deduction rules give a formal proofIntroductory Natural Deduction QuestionGiven P ∨ ¬ P prove (P → Q) → ((¬ P → Q) → Q) by natural deductionHow to construct a counter-model of □P --> □◊P in T and K?Natural deduction proof help!Any solution to prove (∀x)(∃y)(Fx & Gy) ⊢ (∃y)(∀x)(Fx & Gy) with natural deduction?How would i go about using natural deduction to prove this argument is valid?Axiomatic proof of ⊢ □P → □◇□P in S4S5 proof of ⊢◻(◻P→◻Q)∨◻(◻Q→◻P)













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It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.










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  • I made an edit which you may roll back or continue editing. Welcome!

    – Frank Hubeny
    8 hours ago











  • It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.

    – Graham Kemp
    2 hours ago
















1















It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.










share|improve this question









New contributor



zzzz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.




















  • I made an edit which you may roll back or continue editing. Welcome!

    – Frank Hubeny
    8 hours ago











  • It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.

    – Graham Kemp
    2 hours ago














1












1








1








It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.










share|improve this question









New contributor



zzzz is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.











It feels intuitively correct, but I cannot work out how to prove it. I would appreciate any help.







logic deduction






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share|improve this question









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edited 8 hours ago









Frank Hubeny

11.7k51564




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asked 8 hours ago









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  • I made an edit which you may roll back or continue editing. Welcome!

    – Frank Hubeny
    8 hours ago











  • It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.

    – Graham Kemp
    2 hours ago



















  • I made an edit which you may roll back or continue editing. Welcome!

    – Frank Hubeny
    8 hours ago











  • It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.

    – Graham Kemp
    2 hours ago

















I made an edit which you may roll back or continue editing. Welcome!

– Frank Hubeny
8 hours ago





I made an edit which you may roll back or continue editing. Welcome!

– Frank Hubeny
8 hours ago













It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.

– Graham Kemp
2 hours ago





It is not intuitively correct; Two predicates not being equivalent does not prohibit one from implying the other.

– Graham Kemp
2 hours ago










2 Answers
2






active

oldest

votes


















2














The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:



enter image description here



If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.



Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.





Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html






share|improve this answer































    2














    You can't derive ~(A→B) from ~(A↔B).



    Consider:



    A = I'm in Paris.
    B = I'm in France.



    ~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).






    share|improve this answer
























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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      2














      The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:



      enter image description here



      If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.



      Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.





      Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html






      share|improve this answer




























        2














        The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:



        enter image description here



        If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.



        Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.





        Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html






        share|improve this answer


























          2












          2








          2







          The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:



          enter image description here



          If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.



          Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.





          Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html






          share|improve this answer













          The following truth table shows that ~(A↔B) → ~(A→B) is not a tautology:



          enter image description here



          If A is False and B is True then the antecedent is True but the consequent is False making the conditional False.



          Because the truth table does not show a tautology, one should not be able to derive a natural deduction proof of the result.





          Michael Rieppel. Truth Table Generator. https://mrieppel.net/prog/truthtable.html







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered 8 hours ago









          Frank HubenyFrank Hubeny

          11.7k51564




          11.7k51564























              2














              You can't derive ~(A→B) from ~(A↔B).



              Consider:



              A = I'm in Paris.
              B = I'm in France.



              ~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).






              share|improve this answer




























                2














                You can't derive ~(A→B) from ~(A↔B).



                Consider:



                A = I'm in Paris.
                B = I'm in France.



                ~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).






                share|improve this answer


























                  2












                  2








                  2







                  You can't derive ~(A→B) from ~(A↔B).



                  Consider:



                  A = I'm in Paris.
                  B = I'm in France.



                  ~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).






                  share|improve this answer













                  You can't derive ~(A→B) from ~(A↔B).



                  Consider:



                  A = I'm in Paris.
                  B = I'm in France.



                  ~(A↔B) is true, because being in Paris is not equivalent to being in France (I could be in France but not in Paris). But ~(A→B) is false, because if I'm in Paris then necessarily I'm in France. So you can't derive ~(A→B) from ~(A↔B).







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered 5 hours ago









                  EliranEliran

                  4,52231435




                  4,52231435






















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