proof for relational predicate logic











up vote
3
down vote

favorite












I have been working on this problem for over an hour and I think I have simply missed something. I need some help. The rules I am allowed to use are the Basic Inference rules (MP, MT, HS, Simp, Conj, DS, Add, CD), the Replacement Rules (DN, Comm, Assoc, Dup, DeM, BE, Contrap, CE, Exp, Dist.), CP, IP, and EI, UI, EG, UG.



The problem is:




  1. ¬(∃x)(Axa ∧ ~Bxb)

  2. ¬(∃x)(Cxc ∧ Cbx)

  3. (∀x)(Bex → Cxf)


/∴ ¬(Aea ∧ Cfc)



Any help would be appreciated. Thanks.



Edit: Fixed the format of the question. The system is quantificational predicate logic. I've never called it anything else.










share|improve this question




















  • 1




    This formula is unreadable. Please reformat it with ∀, ∃ for quantifiers, ∨, ∧, →, ¬ for connectives (you can copy symbols from here), and predicates as P(x,y,z). What > and // are I can not even guess. "Basic Inference rules" and "Replacement Rules" are specific to your book and not standard terminology, so best provide a reference and what type of system it is using (natural deduction, sequent calculus, something else?).
    – Conifold
    Dec 4 at 0:46








  • 1




    I formatted the premises and conclusion to use Conifold's symbols. You may roll this back or continue editing if I got it wrong.
    – Frank Hubeny
    Dec 4 at 1:32










  • @fantasticorangina Have the answers been of any help ?
    – Graham Kemp
    Dec 5 at 0:45










  • @fantasticorangina Is that a no?
    – Graham Kemp
    Dec 6 at 23:22















up vote
3
down vote

favorite












I have been working on this problem for over an hour and I think I have simply missed something. I need some help. The rules I am allowed to use are the Basic Inference rules (MP, MT, HS, Simp, Conj, DS, Add, CD), the Replacement Rules (DN, Comm, Assoc, Dup, DeM, BE, Contrap, CE, Exp, Dist.), CP, IP, and EI, UI, EG, UG.



The problem is:




  1. ¬(∃x)(Axa ∧ ~Bxb)

  2. ¬(∃x)(Cxc ∧ Cbx)

  3. (∀x)(Bex → Cxf)


/∴ ¬(Aea ∧ Cfc)



Any help would be appreciated. Thanks.



Edit: Fixed the format of the question. The system is quantificational predicate logic. I've never called it anything else.










share|improve this question




















  • 1




    This formula is unreadable. Please reformat it with ∀, ∃ for quantifiers, ∨, ∧, →, ¬ for connectives (you can copy symbols from here), and predicates as P(x,y,z). What > and // are I can not even guess. "Basic Inference rules" and "Replacement Rules" are specific to your book and not standard terminology, so best provide a reference and what type of system it is using (natural deduction, sequent calculus, something else?).
    – Conifold
    Dec 4 at 0:46








  • 1




    I formatted the premises and conclusion to use Conifold's symbols. You may roll this back or continue editing if I got it wrong.
    – Frank Hubeny
    Dec 4 at 1:32










  • @fantasticorangina Have the answers been of any help ?
    – Graham Kemp
    Dec 5 at 0:45










  • @fantasticorangina Is that a no?
    – Graham Kemp
    Dec 6 at 23:22













up vote
3
down vote

favorite









up vote
3
down vote

favorite











I have been working on this problem for over an hour and I think I have simply missed something. I need some help. The rules I am allowed to use are the Basic Inference rules (MP, MT, HS, Simp, Conj, DS, Add, CD), the Replacement Rules (DN, Comm, Assoc, Dup, DeM, BE, Contrap, CE, Exp, Dist.), CP, IP, and EI, UI, EG, UG.



The problem is:




  1. ¬(∃x)(Axa ∧ ~Bxb)

  2. ¬(∃x)(Cxc ∧ Cbx)

  3. (∀x)(Bex → Cxf)


/∴ ¬(Aea ∧ Cfc)



Any help would be appreciated. Thanks.



Edit: Fixed the format of the question. The system is quantificational predicate logic. I've never called it anything else.










share|improve this question















I have been working on this problem for over an hour and I think I have simply missed something. I need some help. The rules I am allowed to use are the Basic Inference rules (MP, MT, HS, Simp, Conj, DS, Add, CD), the Replacement Rules (DN, Comm, Assoc, Dup, DeM, BE, Contrap, CE, Exp, Dist.), CP, IP, and EI, UI, EG, UG.



The problem is:




  1. ¬(∃x)(Axa ∧ ~Bxb)

  2. ¬(∃x)(Cxc ∧ Cbx)

  3. (∀x)(Bex → Cxf)


/∴ ¬(Aea ∧ Cfc)



Any help would be appreciated. Thanks.



Edit: Fixed the format of the question. The system is quantificational predicate logic. I've never called it anything else.







logic symbolic-logic






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 4 at 1:31









Frank Hubeny

6,31451344




6,31451344










asked Dec 4 at 0:29









fantasticorangina

192




192








  • 1




    This formula is unreadable. Please reformat it with ∀, ∃ for quantifiers, ∨, ∧, →, ¬ for connectives (you can copy symbols from here), and predicates as P(x,y,z). What > and // are I can not even guess. "Basic Inference rules" and "Replacement Rules" are specific to your book and not standard terminology, so best provide a reference and what type of system it is using (natural deduction, sequent calculus, something else?).
    – Conifold
    Dec 4 at 0:46








  • 1




    I formatted the premises and conclusion to use Conifold's symbols. You may roll this back or continue editing if I got it wrong.
    – Frank Hubeny
    Dec 4 at 1:32










  • @fantasticorangina Have the answers been of any help ?
    – Graham Kemp
    Dec 5 at 0:45










  • @fantasticorangina Is that a no?
    – Graham Kemp
    Dec 6 at 23:22














  • 1




    This formula is unreadable. Please reformat it with ∀, ∃ for quantifiers, ∨, ∧, →, ¬ for connectives (you can copy symbols from here), and predicates as P(x,y,z). What > and // are I can not even guess. "Basic Inference rules" and "Replacement Rules" are specific to your book and not standard terminology, so best provide a reference and what type of system it is using (natural deduction, sequent calculus, something else?).
    – Conifold
    Dec 4 at 0:46








  • 1




    I formatted the premises and conclusion to use Conifold's symbols. You may roll this back or continue editing if I got it wrong.
    – Frank Hubeny
    Dec 4 at 1:32










  • @fantasticorangina Have the answers been of any help ?
    – Graham Kemp
    Dec 5 at 0:45










  • @fantasticorangina Is that a no?
    – Graham Kemp
    Dec 6 at 23:22








1




1




This formula is unreadable. Please reformat it with ∀, ∃ for quantifiers, ∨, ∧, →, ¬ for connectives (you can copy symbols from here), and predicates as P(x,y,z). What > and // are I can not even guess. "Basic Inference rules" and "Replacement Rules" are specific to your book and not standard terminology, so best provide a reference and what type of system it is using (natural deduction, sequent calculus, something else?).
– Conifold
Dec 4 at 0:46






This formula is unreadable. Please reformat it with ∀, ∃ for quantifiers, ∨, ∧, →, ¬ for connectives (you can copy symbols from here), and predicates as P(x,y,z). What > and // are I can not even guess. "Basic Inference rules" and "Replacement Rules" are specific to your book and not standard terminology, so best provide a reference and what type of system it is using (natural deduction, sequent calculus, something else?).
– Conifold
Dec 4 at 0:46






1




1




I formatted the premises and conclusion to use Conifold's symbols. You may roll this back or continue editing if I got it wrong.
– Frank Hubeny
Dec 4 at 1:32




I formatted the premises and conclusion to use Conifold's symbols. You may roll this back or continue editing if I got it wrong.
– Frank Hubeny
Dec 4 at 1:32












@fantasticorangina Have the answers been of any help ?
– Graham Kemp
Dec 5 at 0:45




@fantasticorangina Have the answers been of any help ?
– Graham Kemp
Dec 5 at 0:45












@fantasticorangina Is that a no?
– Graham Kemp
Dec 6 at 23:22




@fantasticorangina Is that a no?
– Graham Kemp
Dec 6 at 23:22










2 Answers
2






active

oldest

votes

















up vote
2
down vote













It is useful to have a proof checker to aid learning how to use natural deduction. I have linked to one below in the references.



Using that proof checker and the rules described in forallx I was able to prove the result in 22 lines which included 3 lines for the premises and 1 line for the goal.



Although I don't see it listed I assume you have the change of quantifier replacement rule. If not a derivation is in forallx on pages 260-1. Use that to change the first two premises from "~(∃x)" to "(∀x)~".



Next eliminate the universal quantifier by assigning a different name to the variable "x" in each premise. You should chose these names wisely. Look at the goal to try to pick names that will help you reach the goal.



Then use De Morgan rules to transform the lines with a negation in front of the conjunction to a disjunction of negations.



After that preparatory work, I derived something like the following line: "¬Aea ∨ ¬¬Beb". I used disjunction elimination by considering both cases. I wanted to reach the following: "¬Aea ∨ ¬Cfc". Now that is not quite the desired result but with a use of the De Morgan rules I could convert that into the final goal: "¬(Aea ∧ Cfc)".





References



Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






share|improve this answer




























    up vote
    2
    down vote













    Use an Indirect Proof.



    Begin by assuming (Aea ∧ Cfc), use the first premise to derive Beb, the third to derive Cbf, and the second to derive a contradiction.



    Some of your subproofs will likewise be Indirect Proofs; using existential generalisation to derive their contradictions.



    Assuming, that the thing you call IP operates as expected (derive a contradiction from an assumption, thereby deducing that the negation of the assumption holds).






    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








      up vote
      2
      down vote













      It is useful to have a proof checker to aid learning how to use natural deduction. I have linked to one below in the references.



      Using that proof checker and the rules described in forallx I was able to prove the result in 22 lines which included 3 lines for the premises and 1 line for the goal.



      Although I don't see it listed I assume you have the change of quantifier replacement rule. If not a derivation is in forallx on pages 260-1. Use that to change the first two premises from "~(∃x)" to "(∀x)~".



      Next eliminate the universal quantifier by assigning a different name to the variable "x" in each premise. You should chose these names wisely. Look at the goal to try to pick names that will help you reach the goal.



      Then use De Morgan rules to transform the lines with a negation in front of the conjunction to a disjunction of negations.



      After that preparatory work, I derived something like the following line: "¬Aea ∨ ¬¬Beb". I used disjunction elimination by considering both cases. I wanted to reach the following: "¬Aea ∨ ¬Cfc". Now that is not quite the desired result but with a use of the De Morgan rules I could convert that into the final goal: "¬(Aea ∧ Cfc)".





      References



      Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



      P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






      share|improve this answer

























        up vote
        2
        down vote













        It is useful to have a proof checker to aid learning how to use natural deduction. I have linked to one below in the references.



        Using that proof checker and the rules described in forallx I was able to prove the result in 22 lines which included 3 lines for the premises and 1 line for the goal.



        Although I don't see it listed I assume you have the change of quantifier replacement rule. If not a derivation is in forallx on pages 260-1. Use that to change the first two premises from "~(∃x)" to "(∀x)~".



        Next eliminate the universal quantifier by assigning a different name to the variable "x" in each premise. You should chose these names wisely. Look at the goal to try to pick names that will help you reach the goal.



        Then use De Morgan rules to transform the lines with a negation in front of the conjunction to a disjunction of negations.



        After that preparatory work, I derived something like the following line: "¬Aea ∨ ¬¬Beb". I used disjunction elimination by considering both cases. I wanted to reach the following: "¬Aea ∨ ¬Cfc". Now that is not quite the desired result but with a use of the De Morgan rules I could convert that into the final goal: "¬(Aea ∧ Cfc)".





        References



        Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



        P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






        share|improve this answer























          up vote
          2
          down vote










          up vote
          2
          down vote









          It is useful to have a proof checker to aid learning how to use natural deduction. I have linked to one below in the references.



          Using that proof checker and the rules described in forallx I was able to prove the result in 22 lines which included 3 lines for the premises and 1 line for the goal.



          Although I don't see it listed I assume you have the change of quantifier replacement rule. If not a derivation is in forallx on pages 260-1. Use that to change the first two premises from "~(∃x)" to "(∀x)~".



          Next eliminate the universal quantifier by assigning a different name to the variable "x" in each premise. You should chose these names wisely. Look at the goal to try to pick names that will help you reach the goal.



          Then use De Morgan rules to transform the lines with a negation in front of the conjunction to a disjunction of negations.



          After that preparatory work, I derived something like the following line: "¬Aea ∨ ¬¬Beb". I used disjunction elimination by considering both cases. I wanted to reach the following: "¬Aea ∨ ¬Cfc". Now that is not quite the desired result but with a use of the De Morgan rules I could convert that into the final goal: "¬(Aea ∧ Cfc)".





          References



          Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



          P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/






          share|improve this answer












          It is useful to have a proof checker to aid learning how to use natural deduction. I have linked to one below in the references.



          Using that proof checker and the rules described in forallx I was able to prove the result in 22 lines which included 3 lines for the premises and 1 line for the goal.



          Although I don't see it listed I assume you have the change of quantifier replacement rule. If not a derivation is in forallx on pages 260-1. Use that to change the first two premises from "~(∃x)" to "(∀x)~".



          Next eliminate the universal quantifier by assigning a different name to the variable "x" in each premise. You should chose these names wisely. Look at the goal to try to pick names that will help you reach the goal.



          Then use De Morgan rules to transform the lines with a negation in front of the conjunction to a disjunction of negations.



          After that preparatory work, I derived something like the following line: "¬Aea ∨ ¬¬Beb". I used disjunction elimination by considering both cases. I wanted to reach the following: "¬Aea ∨ ¬Cfc". Now that is not quite the desired result but with a use of the De Morgan rules I could convert that into the final goal: "¬(Aea ∧ Cfc)".





          References



          Kevin Klement's JavaScript/PHP Fitch-style natural deduction proof editor and checker http://proofs.openlogicproject.org/



          P. D. Magnus, Tim Button with additions by J. Robert Loftis remixed and revised by Aaron Thomas-Bolduc, Richard Zach, forallx Calgary Remix: An Introduction to Formal Logic, Winter 2018. http://forallx.openlogicproject.org/







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 4 at 1:32









          Frank Hubeny

          6,31451344




          6,31451344






















              up vote
              2
              down vote













              Use an Indirect Proof.



              Begin by assuming (Aea ∧ Cfc), use the first premise to derive Beb, the third to derive Cbf, and the second to derive a contradiction.



              Some of your subproofs will likewise be Indirect Proofs; using existential generalisation to derive their contradictions.



              Assuming, that the thing you call IP operates as expected (derive a contradiction from an assumption, thereby deducing that the negation of the assumption holds).






              share|improve this answer



























                up vote
                2
                down vote













                Use an Indirect Proof.



                Begin by assuming (Aea ∧ Cfc), use the first premise to derive Beb, the third to derive Cbf, and the second to derive a contradiction.



                Some of your subproofs will likewise be Indirect Proofs; using existential generalisation to derive their contradictions.



                Assuming, that the thing you call IP operates as expected (derive a contradiction from an assumption, thereby deducing that the negation of the assumption holds).






                share|improve this answer

























                  up vote
                  2
                  down vote










                  up vote
                  2
                  down vote









                  Use an Indirect Proof.



                  Begin by assuming (Aea ∧ Cfc), use the first premise to derive Beb, the third to derive Cbf, and the second to derive a contradiction.



                  Some of your subproofs will likewise be Indirect Proofs; using existential generalisation to derive their contradictions.



                  Assuming, that the thing you call IP operates as expected (derive a contradiction from an assumption, thereby deducing that the negation of the assumption holds).






                  share|improve this answer














                  Use an Indirect Proof.



                  Begin by assuming (Aea ∧ Cfc), use the first premise to derive Beb, the third to derive Cbf, and the second to derive a contradiction.



                  Some of your subproofs will likewise be Indirect Proofs; using existential generalisation to derive their contradictions.



                  Assuming, that the thing you call IP operates as expected (derive a contradiction from an assumption, thereby deducing that the negation of the assumption holds).







                  share|improve this answer














                  share|improve this answer



                  share|improve this answer








                  edited Dec 4 at 3:17

























                  answered Dec 4 at 3:12









                  Graham Kemp

                  84918




                  84918






























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