How to Prove P(a) → ∀x(P(x) ∨ ¬(x = a)) using Natural DeductionUsing natural deduction rules give a formal proofIntroductory Natural Deduction QuestionProve A ∨ D from A ∨ (B ∧ C) and (¬ B ∨ ¬ C) ∨ D ( LPL Q6.26) without using --> or material implicationGiven P ∨ ¬ P prove (P → Q) → ((¬ P → Q) → Q) by natural deductionHow to prove ¬(p→q) ⊢ p &¬qDoes anyone have a proof checker they prefer using for modal logic?How do you prove law of excluded middle using tertium non datur?How to prove : (( P → Q ) ∨ ( Q → R )) by natural deductionHow to prove ‘∃xP(x)’ from ‘¬∀x(P(x)→Q(x))’How would i go about using natural deduction to prove this argument is valid?

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

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


Using natural deduction rules give a formal proofIntroductory Natural Deduction QuestionProve A ∨ D from A ∨ (B ∧ C) and (¬ B ∨ ¬ C) ∨ D ( LPL Q6.26) without using --> or material implicationGiven P ∨ ¬ P prove (P → Q) → ((¬ P → Q) → Q) by natural deductionHow to prove ¬(p→q) ⊢ p &¬qDoes anyone have a proof checker they prefer using for modal logic?How do you prove law of excluded middle using tertium non datur?How to prove : (( P → Q ) ∨ ( Q → R )) by natural deductionHow to prove ‘∃xP(x)’ from ‘¬∀x(P(x)→Q(x))’How would i go about using natural deduction to prove this argument is valid?













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






      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.











      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




      share|improve this question






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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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      asked 3 hours ago









      Moey mnmMoey mnm

      16




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      New contributor





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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!






          share|improve this answer























          • Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org

            – Frank Hubeny
            17 mins ago











          Your Answer








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          1 Answer
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          1 Answer
          1






          active

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          active

          oldest

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          active

          oldest

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          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























          • Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org

            – Frank Hubeny
            17 mins ago















          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























          • Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org

            – Frank Hubeny
            17 mins ago













          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












          share|improve this answer



          share|improve this answer










          answered 1 hour ago









          AdamAdam

          4358




          4358












          • Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org

            – Frank Hubeny
            17 mins ago

















          • Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org

            – Frank Hubeny
            17 mins ago
















          Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org

          – Frank Hubeny
          17 mins ago





          Just as confirmation, your suggestion for how to proceed worked using the following proof checker: proofs.openlogicproject.org

          – Frank Hubeny
          17 mins ago










          Moey mnm is a new contributor. Be nice, and check out our Code of Conduct.









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