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Give a categorical proof of p -> [q -> q]



The 2019 Stack Overflow Developer Survey Results Are InFalse(ified) AxiomsHilbert calculus: Proof that every provable formula has a proofHelp with axiom in propositional logicWhy does adding material implication as an axiom to propositional calculus make every formula provable?$S5$ proof of $diamond square pto square p$Can someone break this Propositional logic formulae down for me via truth tableProof of $(neg A supset A) supset A$Can't seem to show that (P∧R)⇒¬(Q∧¬(P∧R)) is a tautologyUnderstanding of Material Implication$(p wedge q) wedge p$ convert to CNF










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I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.










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    0












    $begingroup$


    I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.










    share|cite|improve this question











    $endgroup$














      0












      0








      0





      $begingroup$


      I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.










      share|cite|improve this question











      $endgroup$




      I am doing Fitch's Exercises of Symbolic Logic, Chapter 1. This is the first exercise. We have so far axioms such as the distributivity axiom, the axiom of conditioned repetition, the transitivity of implication. I really don't know how to show that this is a theorem.







      logic propositional-calculus






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      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Mar 23 at 4:27









      MJD

      47.8k29216397




      47.8k29216397










      asked Mar 23 at 3:59









      simulacrasimulacra

      463




      463




















          2 Answers
          2






          active

          oldest

          votes


















          2












          $begingroup$

          thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)



          1) q->(q->q) axiom of conditioned repetition
          2) q->(q->q)->q axiom of conditioned repetition
          3) (q->(q->q))->(q->q) by 2 and distributivity
          4) q->q by 1, 3, modus ponies
          5) p->(q->q) axiom of conditioned repetition






          share|cite|improve this answer









          $endgroup$












          • $begingroup$
            modus ponies? $ddotsmile$
            $endgroup$
            – Graham Kemp
            Mar 23 at 14:01


















          0












          $begingroup$

          Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.






          share|cite|improve this answer











          $endgroup$













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            2 Answers
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            2 Answers
            2






            active

            oldest

            votes









            active

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            active

            oldest

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            2












            $begingroup$

            thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)



            1) q->(q->q) axiom of conditioned repetition
            2) q->(q->q)->q axiom of conditioned repetition
            3) (q->(q->q))->(q->q) by 2 and distributivity
            4) q->q by 1, 3, modus ponies
            5) p->(q->q) axiom of conditioned repetition






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              modus ponies? $ddotsmile$
              $endgroup$
              – Graham Kemp
              Mar 23 at 14:01















            2












            $begingroup$

            thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)



            1) q->(q->q) axiom of conditioned repetition
            2) q->(q->q)->q axiom of conditioned repetition
            3) (q->(q->q))->(q->q) by 2 and distributivity
            4) q->q by 1, 3, modus ponies
            5) p->(q->q) axiom of conditioned repetition






            share|cite|improve this answer









            $endgroup$












            • $begingroup$
              modus ponies? $ddotsmile$
              $endgroup$
              – Graham Kemp
              Mar 23 at 14:01













            2












            2








            2





            $begingroup$

            thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)



            1) q->(q->q) axiom of conditioned repetition
            2) q->(q->q)->q axiom of conditioned repetition
            3) (q->(q->q))->(q->q) by 2 and distributivity
            4) q->q by 1, 3, modus ponies
            5) p->(q->q) axiom of conditioned repetition






            share|cite|improve this answer









            $endgroup$



            thanks! I figured out that the proof looks like this (please note that I prove it like this because this is Chapter 1 and so all tools have not been given yet)



            1) q->(q->q) axiom of conditioned repetition
            2) q->(q->q)->q axiom of conditioned repetition
            3) (q->(q->q))->(q->q) by 2 and distributivity
            4) q->q by 1, 3, modus ponies
            5) p->(q->q) axiom of conditioned repetition







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Mar 23 at 7:09









            simulacrasimulacra

            463




            463











            • $begingroup$
              modus ponies? $ddotsmile$
              $endgroup$
              – Graham Kemp
              Mar 23 at 14:01
















            • $begingroup$
              modus ponies? $ddotsmile$
              $endgroup$
              – Graham Kemp
              Mar 23 at 14:01















            $begingroup$
            modus ponies? $ddotsmile$
            $endgroup$
            – Graham Kemp
            Mar 23 at 14:01




            $begingroup$
            modus ponies? $ddotsmile$
            $endgroup$
            – Graham Kemp
            Mar 23 at 14:01











            0












            $begingroup$

            Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.






            share|cite|improve this answer











            $endgroup$

















              0












              $begingroup$

              Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.






              share|cite|improve this answer











              $endgroup$















                0












                0








                0





                $begingroup$

                Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.






                share|cite|improve this answer











                $endgroup$



                Hint: If you can prove $r$, you can use $rto (pto r)$ to prove $pto r$;.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                answered Mar 23 at 4:23


























                community wiki





                MJD




























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