When does $(square P land square Q) to square (P land Q)$ hold?











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If all axioms of classical propositional calculus hold and we work in modal logic that is at least K (ie. extremely weak), it is trivial to show $square(P land Q) to (square P land square Q)$. What other modality axioms, if any, have to be added for $(square P land square Q) to square (P land Q)$ to hold?










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    If all axioms of classical propositional calculus hold and we work in modal logic that is at least K (ie. extremely weak), it is trivial to show $square(P land Q) to (square P land square Q)$. What other modality axioms, if any, have to be added for $(square P land square Q) to square (P land Q)$ to hold?










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      favorite











      If all axioms of classical propositional calculus hold and we work in modal logic that is at least K (ie. extremely weak), it is trivial to show $square(P land Q) to (square P land square Q)$. What other modality axioms, if any, have to be added for $(square P land square Q) to square (P land Q)$ to hold?










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      If all axioms of classical propositional calculus hold and we work in modal logic that is at least K (ie. extremely weak), it is trivial to show $square(P land Q) to (square P land square Q)$. What other modality axioms, if any, have to be added for $(square P land square Q) to square (P land Q)$ to hold?







      logic propositional-calculus proof-theory modal-logic hilbert-calculus






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      edited Nov 12 at 23:21









      Taroccoesbrocco

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      asked Nov 12 at 17:37









      Michał Zapała

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          You don't need any other axiom, $(square P land square Q) to square (P land Q)$ is derivable in the system $mathbf{K}$, i.e. in the propositional calculus (with modus ponens and closure under substitution) augmented with the necessitation rule (i.e. if $A$ is derivable then $square A$ is derivable) and the distribution axiom (i.e. $square (P to Q) to (square P to square Q)$).



          A formal proof in Hilbert system for $mathbf{K}$ is the following:





          1. $P to (Q to (P land Q)) qquad$ (axiom for conjunction in propositional calculus)


          2. $square (P to (Q to (P land Q))) qquad$ (necessitation rule applied to 1.)


          3. $square (P to (Q to (P land Q))) to (square P to square (Q to (P land Q))) qquad$ (distribution axiom)


          4. $square P to square (Q to (P land Q)) qquad$ (modus ponens of 3. and 2.)


          5. $square (Q to (P land Q)) to (square Q to square (P land Q)) qquad$ (distribution axiom)

          6. $(square P to square(Q to (P land Q)) ) to big( ( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) big) qquad (*)$


          7. $( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) qquad$ (modus ponens of 6. and 4.)



          8. $(square P land square Q) to square(P land Q) qquad$ (modus ponens of 7. and 5.)


          where the formula in ($*$) is derivable because it is an instance of the tautology in propositional $(p to q) to big( (q to (r to s)) to ((p land r) to s)big)$ obtained through the substitution
          begin{equation}begin{cases}
          p mapsto square P \
          q mapsto square (Q to (P land Q)) \
          r mapsto square Q \
          s mapsto square (P land Q)
          end{cases}end{equation}



          Remind that, by the completeness theorem, any tautology in propositional calculus can be derived in a formal proof system such the Hilbert system.



          For a reference, you can see "A New Introduction to Modal Logic" by Hughes and Cresswell, p. 27.






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            You don't need any other axiom, $(square P land square Q) to square (P land Q)$ is derivable in the system $mathbf{K}$, i.e. in the propositional calculus (with modus ponens and closure under substitution) augmented with the necessitation rule (i.e. if $A$ is derivable then $square A$ is derivable) and the distribution axiom (i.e. $square (P to Q) to (square P to square Q)$).



            A formal proof in Hilbert system for $mathbf{K}$ is the following:





            1. $P to (Q to (P land Q)) qquad$ (axiom for conjunction in propositional calculus)


            2. $square (P to (Q to (P land Q))) qquad$ (necessitation rule applied to 1.)


            3. $square (P to (Q to (P land Q))) to (square P to square (Q to (P land Q))) qquad$ (distribution axiom)


            4. $square P to square (Q to (P land Q)) qquad$ (modus ponens of 3. and 2.)


            5. $square (Q to (P land Q)) to (square Q to square (P land Q)) qquad$ (distribution axiom)

            6. $(square P to square(Q to (P land Q)) ) to big( ( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) big) qquad (*)$


            7. $( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) qquad$ (modus ponens of 6. and 4.)



            8. $(square P land square Q) to square(P land Q) qquad$ (modus ponens of 7. and 5.)


            where the formula in ($*$) is derivable because it is an instance of the tautology in propositional $(p to q) to big( (q to (r to s)) to ((p land r) to s)big)$ obtained through the substitution
            begin{equation}begin{cases}
            p mapsto square P \
            q mapsto square (Q to (P land Q)) \
            r mapsto square Q \
            s mapsto square (P land Q)
            end{cases}end{equation}



            Remind that, by the completeness theorem, any tautology in propositional calculus can be derived in a formal proof system such the Hilbert system.



            For a reference, you can see "A New Introduction to Modal Logic" by Hughes and Cresswell, p. 27.






            share|cite|improve this answer



























              up vote
              4
              down vote



              accepted










              You don't need any other axiom, $(square P land square Q) to square (P land Q)$ is derivable in the system $mathbf{K}$, i.e. in the propositional calculus (with modus ponens and closure under substitution) augmented with the necessitation rule (i.e. if $A$ is derivable then $square A$ is derivable) and the distribution axiom (i.e. $square (P to Q) to (square P to square Q)$).



              A formal proof in Hilbert system for $mathbf{K}$ is the following:





              1. $P to (Q to (P land Q)) qquad$ (axiom for conjunction in propositional calculus)


              2. $square (P to (Q to (P land Q))) qquad$ (necessitation rule applied to 1.)


              3. $square (P to (Q to (P land Q))) to (square P to square (Q to (P land Q))) qquad$ (distribution axiom)


              4. $square P to square (Q to (P land Q)) qquad$ (modus ponens of 3. and 2.)


              5. $square (Q to (P land Q)) to (square Q to square (P land Q)) qquad$ (distribution axiom)

              6. $(square P to square(Q to (P land Q)) ) to big( ( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) big) qquad (*)$


              7. $( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) qquad$ (modus ponens of 6. and 4.)



              8. $(square P land square Q) to square(P land Q) qquad$ (modus ponens of 7. and 5.)


              where the formula in ($*$) is derivable because it is an instance of the tautology in propositional $(p to q) to big( (q to (r to s)) to ((p land r) to s)big)$ obtained through the substitution
              begin{equation}begin{cases}
              p mapsto square P \
              q mapsto square (Q to (P land Q)) \
              r mapsto square Q \
              s mapsto square (P land Q)
              end{cases}end{equation}



              Remind that, by the completeness theorem, any tautology in propositional calculus can be derived in a formal proof system such the Hilbert system.



              For a reference, you can see "A New Introduction to Modal Logic" by Hughes and Cresswell, p. 27.






              share|cite|improve this answer

























                up vote
                4
                down vote



                accepted







                up vote
                4
                down vote



                accepted






                You don't need any other axiom, $(square P land square Q) to square (P land Q)$ is derivable in the system $mathbf{K}$, i.e. in the propositional calculus (with modus ponens and closure under substitution) augmented with the necessitation rule (i.e. if $A$ is derivable then $square A$ is derivable) and the distribution axiom (i.e. $square (P to Q) to (square P to square Q)$).



                A formal proof in Hilbert system for $mathbf{K}$ is the following:





                1. $P to (Q to (P land Q)) qquad$ (axiom for conjunction in propositional calculus)


                2. $square (P to (Q to (P land Q))) qquad$ (necessitation rule applied to 1.)


                3. $square (P to (Q to (P land Q))) to (square P to square (Q to (P land Q))) qquad$ (distribution axiom)


                4. $square P to square (Q to (P land Q)) qquad$ (modus ponens of 3. and 2.)


                5. $square (Q to (P land Q)) to (square Q to square (P land Q)) qquad$ (distribution axiom)

                6. $(square P to square(Q to (P land Q)) ) to big( ( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) big) qquad (*)$


                7. $( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) qquad$ (modus ponens of 6. and 4.)



                8. $(square P land square Q) to square(P land Q) qquad$ (modus ponens of 7. and 5.)


                where the formula in ($*$) is derivable because it is an instance of the tautology in propositional $(p to q) to big( (q to (r to s)) to ((p land r) to s)big)$ obtained through the substitution
                begin{equation}begin{cases}
                p mapsto square P \
                q mapsto square (Q to (P land Q)) \
                r mapsto square Q \
                s mapsto square (P land Q)
                end{cases}end{equation}



                Remind that, by the completeness theorem, any tautology in propositional calculus can be derived in a formal proof system such the Hilbert system.



                For a reference, you can see "A New Introduction to Modal Logic" by Hughes and Cresswell, p. 27.






                share|cite|improve this answer














                You don't need any other axiom, $(square P land square Q) to square (P land Q)$ is derivable in the system $mathbf{K}$, i.e. in the propositional calculus (with modus ponens and closure under substitution) augmented with the necessitation rule (i.e. if $A$ is derivable then $square A$ is derivable) and the distribution axiom (i.e. $square (P to Q) to (square P to square Q)$).



                A formal proof in Hilbert system for $mathbf{K}$ is the following:





                1. $P to (Q to (P land Q)) qquad$ (axiom for conjunction in propositional calculus)


                2. $square (P to (Q to (P land Q))) qquad$ (necessitation rule applied to 1.)


                3. $square (P to (Q to (P land Q))) to (square P to square (Q to (P land Q))) qquad$ (distribution axiom)


                4. $square P to square (Q to (P land Q)) qquad$ (modus ponens of 3. and 2.)


                5. $square (Q to (P land Q)) to (square Q to square (P land Q)) qquad$ (distribution axiom)

                6. $(square P to square(Q to (P land Q)) ) to big( ( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) big) qquad (*)$


                7. $( square(Q to (P land Q)) to (square Q to square(P land Q) )) to ((square P land square Q) to square(P land Q) qquad$ (modus ponens of 6. and 4.)



                8. $(square P land square Q) to square(P land Q) qquad$ (modus ponens of 7. and 5.)


                where the formula in ($*$) is derivable because it is an instance of the tautology in propositional $(p to q) to big( (q to (r to s)) to ((p land r) to s)big)$ obtained through the substitution
                begin{equation}begin{cases}
                p mapsto square P \
                q mapsto square (Q to (P land Q)) \
                r mapsto square Q \
                s mapsto square (P land Q)
                end{cases}end{equation}



                Remind that, by the completeness theorem, any tautology in propositional calculus can be derived in a formal proof system such the Hilbert system.



                For a reference, you can see "A New Introduction to Modal Logic" by Hughes and Cresswell, p. 27.







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                edited Nov 12 at 23:22

























                answered Nov 12 at 18:58









                Taroccoesbrocco

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