How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural Deduction












3















How would a formal Fitch proof look like.
I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
I am confused on how to proceed with the proof.
Please advice me on how to go about with this.



Thanks in advance










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    3















    How would a formal Fitch proof look like.
    I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
    I am confused on how to proceed with the proof.
    Please advice me on how to go about with this.



    Thanks in advance










    share|improve this question







    New contributor




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























      3












      3








      3








      How would a formal Fitch proof look like.
      I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
      I am confused on how to proceed with the proof.
      Please advice me on how to go about with this.



      Thanks in advance










      share|improve this question







      New contributor




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












      How would a formal Fitch proof look like.
      I am given P(a) → ∀x(P(x) ∨ ¬(x = a)) to prove using Natural Deduction of predicate logic.
      I am confused on how to proceed with the proof.
      Please advice me on how to go about with this.



      Thanks in advance







      logic proof fitch quantification






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      Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question







      New contributor




      Moey mnm is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|improve this question






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      asked 2 hours ago









      Moey mnmMoey mnm

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          HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).



          The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be stand for the particular such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).



          Hope this helps!






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            1 Answer
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            1 Answer
            1






            active

            oldest

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            active

            oldest

            votes






            active

            oldest

            votes









            2














            HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).



            The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be stand for the particular such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).



            Hope this helps!






            share|improve this answer




























              2














              HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).



              The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be stand for the particular such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).



              Hope this helps!






              share|improve this answer


























                2












                2








                2







                HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).



                The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be stand for the particular such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).



                Hope this helps!






                share|improve this answer













                HINT: I'll sketch the derivation. Since the theorem is a conditional, try using conditional proof/conditional-introduction by assuming P(a) and trying to derive ∀x(P(x) ∨ ¬(x = a)) from it. Here, to derive it, I would try an indirect proof by assuming the negation ¬∀x(P(x) ∨ ¬(x = a)) and trying to derive a contradiction. Use quantifier equivalence rules to get ∃x¬(P(x) ∨ ¬(x = a)).



                The next steps will be a little different depending on your list of rules (quantifier rules typically come with restrictions to ensure the rules are sound, and different texts will use different restrictions). Roughly, we can let y be stand for the particular such that ¬(P(y) ∨ ¬(y = a)). Apply De Morgan's law to get ¬P(y) ∧ (y = a). Since y = a, it must be that ¬P(a), contradicting our assumption that P(a). Hence our contradiction completing the indirect proof of ∀x(P(x) ∨ ¬(x = a)).



                Hope this helps!







                share|improve this answer












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                answered 56 mins ago









                AdamAdam

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