Is $A & B ⊸ A$ derivable?












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Intuitively, the sentence $A & B ⊸ A$ seems to mean "Using a choice between $A$ and $B$, get an $A$." This feels like it should be derivable for any $A$ and $B$, but I haven't found any way to derive it from the definition of $&$. Is it possible to establish this in linear logic? Or, if not, what makes this sentence different from the definition of $&$?






linear-logic







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    1












    $begingroup$


    Intuitively, the sentence $A & B ⊸ A$ seems to mean "Using a choice between $A$ and $B$, get an $A$." This feels like it should be derivable for any $A$ and $B$, but I haven't found any way to derive it from the definition of $&$. Is it possible to establish this in linear logic? Or, if not, what makes this sentence different from the definition of $&$?






    linear-logic







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Intuitively, the sentence $A & B ⊸ A$ seems to mean "Using a choice between $A$ and $B$, get an $A$." This feels like it should be derivable for any $A$ and $B$, but I haven't found any way to derive it from the definition of $&$. Is it possible to establish this in linear logic? Or, if not, what makes this sentence different from the definition of $&$?






      linear-logic







      share|cite|improve this question











      $endgroup$




      Intuitively, the sentence $A & B ⊸ A$ seems to mean "Using a choice between $A$ and $B$, get an $A$." This feels like it should be derivable for any $A$ and $B$, but I haven't found any way to derive it from the definition of $&$. Is it possible to establish this in linear logic? Or, if not, what makes this sentence different from the definition of $&$?






      linear-logic




      linear-logic






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      edited Dec 12 '18 at 21:01







      Woofmao

















      asked Dec 12 '18 at 20:36









      WoofmaoWoofmao

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      166






















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          $begingroup$

          It turns out that $A & B multimap A$ is provable. Here is the proof, or at least my best attempt at writing it:



          $$DeclareMathOperator{par}{unicode{8523}}
          cfrac
          {cfrac
          {cfrac
          {init}
          {vdash A^bot par A}}
          {vdash (A^botoplus B^bot) par A}}
          {vdash A & B multimap A}$$



          Using a similar method, you can also prove $A multimap Aoplus B$.






          share|cite|improve this answer









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            $begingroup$

            It turns out that $A & B multimap A$ is provable. Here is the proof, or at least my best attempt at writing it:



            $$DeclareMathOperator{par}{unicode{8523}}
            cfrac
            {cfrac
            {cfrac
            {init}
            {vdash A^bot par A}}
            {vdash (A^botoplus B^bot) par A}}
            {vdash A & B multimap A}$$



            Using a similar method, you can also prove $A multimap Aoplus B$.






            share|cite|improve this answer









            $endgroup$


















              0












              $begingroup$

              It turns out that $A & B multimap A$ is provable. Here is the proof, or at least my best attempt at writing it:



              $$DeclareMathOperator{par}{unicode{8523}}
              cfrac
              {cfrac
              {cfrac
              {init}
              {vdash A^bot par A}}
              {vdash (A^botoplus B^bot) par A}}
              {vdash A & B multimap A}$$



              Using a similar method, you can also prove $A multimap Aoplus B$.






              share|cite|improve this answer









              $endgroup$
















                0












                0








                0





                $begingroup$

                It turns out that $A & B multimap A$ is provable. Here is the proof, or at least my best attempt at writing it:



                $$DeclareMathOperator{par}{unicode{8523}}
                cfrac
                {cfrac
                {cfrac
                {init}
                {vdash A^bot par A}}
                {vdash (A^botoplus B^bot) par A}}
                {vdash A & B multimap A}$$



                Using a similar method, you can also prove $A multimap Aoplus B$.






                share|cite|improve this answer









                $endgroup$



                It turns out that $A & B multimap A$ is provable. Here is the proof, or at least my best attempt at writing it:



                $$DeclareMathOperator{par}{unicode{8523}}
                cfrac
                {cfrac
                {cfrac
                {init}
                {vdash A^bot par A}}
                {vdash (A^botoplus B^bot) par A}}
                {vdash A & B multimap A}$$



                Using a similar method, you can also prove $A multimap Aoplus B$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 14 '18 at 0:57









                WoofmaoWoofmao

                166




                166






























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