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I'm stuck and having trouble with ¬P ∨ Q Prove: P → Q


Help with simple deductive proofInvalid arguments with true premises and true conclusionIntroductory Natural Deduction QuestionHow to prove the axiom is wrong?What are the important effects of studying logic?If F is a sufficient condition for G, is lacking G a sufficient condition for lacking F?How to prove the tautology ¬(P↔¬P) using Fitch?How do you prove B v A |- A v B?S5 proof of ⊢◻(◻P→◻Q)∨◻(◻Q→◻P)trouble with rules of inference practice problems













0















I am having trouble with this problem as I have just started doing logic. Is this the same as




P → Q

Prove: ¬P ∨ Q




?










share|improve this question









New contributor




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















  • 3





    Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org

    – Frank Hubeny
    Apr 23 at 1:08






  • 1





    Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".

    – Jishin Noben
    2 days ago











  • The two statements should have the same truth values. Would that analysis assist in showing equivalence?

    – Mark Andrews
    2 days ago















0















I am having trouble with this problem as I have just started doing logic. Is this the same as




P → Q

Prove: ¬P ∨ Q




?










share|improve this question









New contributor




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















  • 3





    Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org

    – Frank Hubeny
    Apr 23 at 1:08






  • 1





    Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".

    – Jishin Noben
    2 days ago











  • The two statements should have the same truth values. Would that analysis assist in showing equivalence?

    – Mark Andrews
    2 days ago













0












0








0








I am having trouble with this problem as I have just started doing logic. Is this the same as




P → Q

Prove: ¬P ∨ Q




?










share|improve this question









New contributor




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












I am having trouble with this problem as I have just started doing logic. Is this the same as




P → Q

Prove: ¬P ∨ Q




?







logic






share|improve this question









New contributor




Hamish Docherty 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




Hamish Docherty 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




share|improve this question








edited 2 days ago









Glorfindel

1851210




1851210






New contributor




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









asked Apr 23 at 0:53









Hamish DochertyHamish Docherty

102




102




New contributor




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





New contributor





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






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







  • 3





    Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org

    – Frank Hubeny
    Apr 23 at 1:08






  • 1





    Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".

    – Jishin Noben
    2 days ago











  • The two statements should have the same truth values. Would that analysis assist in showing equivalence?

    – Mark Andrews
    2 days ago












  • 3





    Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org

    – Frank Hubeny
    Apr 23 at 1:08






  • 1





    Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".

    – Jishin Noben
    2 days ago











  • The two statements should have the same truth values. Would that analysis assist in showing equivalence?

    – Mark Andrews
    2 days ago







3




3





Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org

– Frank Hubeny
Apr 23 at 1:08





Which text book are you using? An online proof checker and text book may be helpful as supplementary material: proofs.openlogicproject.org

– Frank Hubeny
Apr 23 at 1:08




1




1





Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".

– Jishin Noben
2 days ago





Welcome to PSE. The answers hint at how to find a proof. But your question seems to be whether proving P → Q from ¬P ∨ Q is the same as proving ¬P ∨ Q from P → Q to which the answer is "No, this is not the same thing, though the proofs might look (structurally or otherwise) similar".

– Jishin Noben
2 days ago













The two statements should have the same truth values. Would that analysis assist in showing equivalence?

– Mark Andrews
2 days ago





The two statements should have the same truth values. Would that analysis assist in showing equivalence?

– Mark Andrews
2 days ago










3 Answers
3






active

oldest

votes


















5














In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
Proof.



This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).



Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.






share|improve this answer






























    2














    In this particular case, the two statements are equivalent:
    (¬P ∨ Q) ⊢ (P → Q) and (P → Q) ⊢ (¬P ∨ Q)
    are both provably true statements, so (¬P ∨ Q) ≡ (P → Q).



    But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q) and (P → Q), we have these two statements:



    • P

    • P ∨ Q

    We can trivially prove that (P ∨ Q) follows from P; this is more or less the definition of the addition rule. But P does not necessarily follow from (P ∨ Q), since (¬P ∧ Q) also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.






    share|improve this answer























    • The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.

      – Logikal
      2 days ago











    • @Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.

      – Ray
      2 days ago



















    0














    If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)



    This shows that the two statements are equivalent.



    However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:



    enter image description here



    See the links below for an explanation of the rules used by this proof.



    The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:



    enter image description here



    Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.




    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























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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      5














      In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
      Proof.



      This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).



      Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.






      share|improve this answer



























        5














        In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
        Proof.



        This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).



        Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.






        share|improve this answer

























          5












          5








          5







          In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
          Proof.



          This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).



          Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.






          share|improve this answer













          In a natural deduction system (if that is what you are using) to prove a conditional, such as is P → Q, you must use a Conditional
          Proof.



          This takes the form of assuming the antecedent (that is P) aiming to derive the consequent (that is Q) through valid inferences (also using the premises; that is ¬P ∨ Q). Then discharging the assumption allow the deduction of the conditional (that is P → Q).



          Now to prove Q from an assumption of P and the premise of ¬P ∨ Q, either use Disjunctive Syllogism, or a Proof by Cases.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Apr 23 at 4:47









          Graham KempGraham Kemp

          1,08919




          1,08919





















              2














              In this particular case, the two statements are equivalent:
              (¬P ∨ Q) ⊢ (P → Q) and (P → Q) ⊢ (¬P ∨ Q)
              are both provably true statements, so (¬P ∨ Q) ≡ (P → Q).



              But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q) and (P → Q), we have these two statements:



              • P

              • P ∨ Q

              We can trivially prove that (P ∨ Q) follows from P; this is more or less the definition of the addition rule. But P does not necessarily follow from (P ∨ Q), since (¬P ∧ Q) also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.






              share|improve this answer























              • The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.

                – Logikal
                2 days ago











              • @Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.

                – Ray
                2 days ago
















              2














              In this particular case, the two statements are equivalent:
              (¬P ∨ Q) ⊢ (P → Q) and (P → Q) ⊢ (¬P ∨ Q)
              are both provably true statements, so (¬P ∨ Q) ≡ (P → Q).



              But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q) and (P → Q), we have these two statements:



              • P

              • P ∨ Q

              We can trivially prove that (P ∨ Q) follows from P; this is more or less the definition of the addition rule. But P does not necessarily follow from (P ∨ Q), since (¬P ∧ Q) also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.






              share|improve this answer























              • The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.

                – Logikal
                2 days ago











              • @Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.

                – Ray
                2 days ago














              2












              2








              2







              In this particular case, the two statements are equivalent:
              (¬P ∨ Q) ⊢ (P → Q) and (P → Q) ⊢ (¬P ∨ Q)
              are both provably true statements, so (¬P ∨ Q) ≡ (P → Q).



              But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q) and (P → Q), we have these two statements:



              • P

              • P ∨ Q

              We can trivially prove that (P ∨ Q) follows from P; this is more or less the definition of the addition rule. But P does not necessarily follow from (P ∨ Q), since (¬P ∧ Q) also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.






              share|improve this answer













              In this particular case, the two statements are equivalent:
              (¬P ∨ Q) ⊢ (P → Q) and (P → Q) ⊢ (¬P ∨ Q)
              are both provably true statements, so (¬P ∨ Q) ≡ (P → Q).



              But in order to prove that equivalence, we need to prove both directions separately. To see why, consider the case where instead of (¬P ∨ Q) and (P → Q), we have these two statements:



              • P

              • P ∨ Q

              We can trivially prove that (P ∨ Q) follows from P; this is more or less the definition of the addition rule. But P does not necessarily follow from (P ∨ Q), since (¬P ∧ Q) also satisfies that clause. We can prove it in one direction, but they are not equivalent statements.







              share|improve this answer












              share|improve this answer



              share|improve this answer










              answered 2 days ago









              RayRay

              27217




              27217












              • The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.

                – Logikal
                2 days ago











              • @Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.

                – Ray
                2 days ago


















              • The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.

                – Logikal
                2 days ago











              • @Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.

                – Ray
                2 days ago

















              The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.

              – Logikal
              2 days ago





              The two statements are logically equivalent using the material implication rule. Draw out a truth table and see for yourself that the tables are 100 percent identical.

              – Logikal
              2 days ago













              @Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.

              – Ray
              2 days ago






              @Logikal I know they are equivalent. I said they are equivalent. But the OP isn't asking if they're equivalent; they are asking if, having already proven A ⊢ B, you must still prove B ⊢ A in order to establish that an equivalence exists. The answer to that is "yes". The fact that this particular equivalence has already been proven doesn't change that.

              – Ray
              2 days ago












              0














              If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)



              This shows that the two statements are equivalent.



              However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:



              enter image description here



              See the links below for an explanation of the rules used by this proof.



              The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:



              enter image description here



              Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.




              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



























                0














                If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)



                This shows that the two statements are equivalent.



                However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:



                enter image description here



                See the links below for an explanation of the rules used by this proof.



                The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:



                enter image description here



                Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.




                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

























                  0












                  0








                  0







                  If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)



                  This shows that the two statements are equivalent.



                  However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:



                  enter image description here



                  See the links below for an explanation of the rules used by this proof.



                  The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:



                  enter image description here



                  Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.




                  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













                  If one uses a truth table generator one can show that (¬P ∨ Q) ↔ (P → Q). To see this, insert the following input into the Stanford Truth Table Tool: (~P or Q)<=>(P=>Q)



                  This shows that the two statements are equivalent.



                  However, the question asks one to prove P → Q given the premise ¬P ∨ Q. Here is how one might do this using a natural deduction proof checker:



                  enter image description here



                  See the links below for an explanation of the rules used by this proof.



                  The proof would not be the same if we wanted to prove ¬P ∨ Q given the premise P → Q. That proof would look like this:



                  enter image description here



                  Although the statements are equivalent based on a truth table generator, the proofs from one to the other may be different depending on which is the premise and which the conclusion.




                  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 yesterday









                  Frank HubenyFrank Hubeny

                  10.8k51559




                  10.8k51559




















                      Hamish Docherty is a new contributor. Be nice, and check out our Code of Conduct.









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                      Hamish Docherty is a new contributor. Be nice, and check out our Code of Conduct.














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