2 simple Formal Fitch Proofs Help!

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











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I'm having difficulty proving these. I've been stuck on them for nearly 4 hours. They seem obvious, but I can't figure how to set up FORMAL proofs for them. Could anyone give me at least clues how to start them?



  1. ¬(P∧¬Q) from the premise P→Q;


  2. ¬Q→(R→P) from the premise P∨Q.


Thanks!






proof symbolic-logic fitch







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  • I made some edits which you may roll back or continue editing. Welcome to this SE! You might find some hints by searching on the fitch or symbolic-logic tags.
    – Frank Hubeny
    3 hours ago










  • @FrankHubeny thank you!
    – Yanik
    2 hours ago










  • it would be nice if you separated them into separate questions as there are two different things. Also, what system of proof are you using?
    – virmaior
    2 hours ago














up vote
2
down vote

favorite












I'm having difficulty proving these. I've been stuck on them for nearly 4 hours. They seem obvious, but I can't figure how to set up FORMAL proofs for them. Could anyone give me at least clues how to start them?



  1. ¬(P∧¬Q) from the premise P→Q;


  2. ¬Q→(R→P) from the premise P∨Q.


Thanks!






proof symbolic-logic fitch







share|improve this question









New contributor




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



















  • I made some edits which you may roll back or continue editing. Welcome to this SE! You might find some hints by searching on the fitch or symbolic-logic tags.
    – Frank Hubeny
    3 hours ago










  • @FrankHubeny thank you!
    – Yanik
    2 hours ago










  • it would be nice if you separated them into separate questions as there are two different things. Also, what system of proof are you using?
    – virmaior
    2 hours ago












up vote
2
down vote

favorite









up vote
2
down vote

favorite











I'm having difficulty proving these. I've been stuck on them for nearly 4 hours. They seem obvious, but I can't figure how to set up FORMAL proofs for them. Could anyone give me at least clues how to start them?



  1. ¬(P∧¬Q) from the premise P→Q;


  2. ¬Q→(R→P) from the premise P∨Q.


Thanks!






proof symbolic-logic fitch







share|improve this question









New contributor




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











I'm having difficulty proving these. I've been stuck on them for nearly 4 hours. They seem obvious, but I can't figure how to set up FORMAL proofs for them. Could anyone give me at least clues how to start them?



  1. ¬(P∧¬Q) from the premise P→Q;


  2. ¬Q→(R→P) from the premise P∨Q.


Thanks!






proof symbolic-logic fitch




proof symbolic-logic fitch






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edited 3 hours ago









Frank Hubeny

3,9943837




3,9943837






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asked 3 hours ago









Yanik

132




132




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  • I made some edits which you may roll back or continue editing. Welcome to this SE! You might find some hints by searching on the fitch or symbolic-logic tags.
    – Frank Hubeny
    3 hours ago










  • @FrankHubeny thank you!
    – Yanik
    2 hours ago










  • it would be nice if you separated them into separate questions as there are two different things. Also, what system of proof are you using?
    – virmaior
    2 hours ago
















  • I made some edits which you may roll back or continue editing. Welcome to this SE! You might find some hints by searching on the fitch or symbolic-logic tags.
    – Frank Hubeny
    3 hours ago










  • @FrankHubeny thank you!
    – Yanik
    2 hours ago










  • it would be nice if you separated them into separate questions as there are two different things. Also, what system of proof are you using?
    – virmaior
    2 hours ago















I made some edits which you may roll back or continue editing. Welcome to this SE! You might find some hints by searching on the fitch or symbolic-logic tags.
– Frank Hubeny
3 hours ago




I made some edits which you may roll back or continue editing. Welcome to this SE! You might find some hints by searching on the fitch or symbolic-logic tags.
– Frank Hubeny
3 hours ago












@FrankHubeny thank you!
– Yanik
2 hours ago




@FrankHubeny thank you!
– Yanik
2 hours ago












it would be nice if you separated them into separate questions as there are two different things. Also, what system of proof are you using?
– virmaior
2 hours ago




it would be nice if you separated them into separate questions as there are two different things. Also, what system of proof are you using?
– virmaior
2 hours ago










2 Answers
2






active

oldest

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up vote
1
down vote



accepted











¬(P∧¬Q) from the premise P→Q




Well, I know you stated your main issue is coming up with formal proofs (by which I'm going to assume you mean something like natural deduction) but let's just get started with the simplest conceptual level.



  1. P→Q

this means that either P is false or Q is true based on the truth table that defines →.



Using a truth table, we can then check to see if the two are equivalent:



 P | Q | P → Q | ¬Q | P ∧ ¬Q | ¬(P∧¬Q)
 T T T F F T
 T F F T T F
 F T T F F T
 F F T T F T


So we've shown they are equivalent via truth tables.



Now, how do we move that into the realm of natural deduction?



The two basic options are:



  1. use inference rules

  2. do a sub proof -- either to get a contradiction or supply a conditional proof

Since our conclusion has a negation, let's go with a subproof by contradiciton.



  1. P→Q A

  2. | P∧¬Q A

  3. | P ∧E 2

  4. | Q MP 1,3

  5. | ¬Q ∧E 2

  6. | ⊥ contradiction on 4,5

  7. ¬(P∧¬Q) RAA 2-6

enter image description here



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/



(references copied from Frank Hubeny)






share|improve this answer





























    up vote
    1
    down vote













    I agree with virmaior's proof. Since I started these as well I figured I might as well leave them here since they contain a different presentation.



    Here are two proofs using Klement's proof checker. The rules you may have to use may be different.



    enter image description here



    The proof uses conjunction elimination (∧E), conditional elimination (→E), contradiction introduction (⊥I) and negation introduction (¬I).



    To avoid the not ("¬") in front of "P∧¬Q" I started an assumption on line 2 assuming "P∧¬Q" hoping to derive a contradiction which I did on line 6. To get to that contradiction I split the conjunction into the two conjuncts, "P" and "¬Q" on lines 3 and 4. Then having "P" on line 3 and the conditional on line 1, I could put "Q" on line 5. Since lines 4 and 5 contradicted each other I could introduce a contradiction on line 6. This allowed me to discharge the assumption on line 2 and introduce the negative in from of "P∧¬Q".



    For the other one, try something like this:



    enter image description here



    The proof uses contradiction introduction (⊥I), explosion (X), disjunction elimination (∨E) and conditional introduction (→I).



    Since the goal is a conditional "¬Q → (R→P)", I assume "¬Q" on line 2 and hope to derive "R→P". Since what I want to derive is a conditional as well, I assume the premise of that conditional which is "R" on line 3. At this point I need to derive "P".



    There is a "P" on line 1, but that is a disjunction. In order to get that "P", I have to deal with the "Q" as well. The first case is easy. I have to start a subproof with just "P" as the assumption and the conclusion. This subproof is just one item, line 4. Having taken care of "P", I then have to take care of "Q". Fortunately line 2 and 5 contradict each other which allows me to introduce a contradiction on line 6. Having a contradiction I can use explosion to get anything whatsoever. I need "P", so I put "P" on line 7. This allows me to discharge both the "P" assumption on line 4 and the "Q" assumption on line 5 by eliminating the disjunction of line 1 on line 8.



    At this point note that I have a set up with "R" on line 3 and "P" on line 8. That means I can discharge the assumption of "R" on line 3 and put "R→P" on line 9. That allows me to discharge the assumption "¬Q" on line 2 and get the final result on line 10.



    You can use this proof checker online to help clarify the details yourself. The accompanying text forall x is available for further information. What you are required to use may be different. You will likely need to understand the above proofs well enough to present it in the format you need.



    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






















    • Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful!
      – Yanik
      33 mins ago










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    2 Answers
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    2 Answers
    2






    active

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    active

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    up vote
    1
    down vote



    accepted











    ¬(P∧¬Q) from the premise P→Q




    Well, I know you stated your main issue is coming up with formal proofs (by which I'm going to assume you mean something like natural deduction) but let's just get started with the simplest conceptual level.



    1. P→Q

    this means that either P is false or Q is true based on the truth table that defines →.



    Using a truth table, we can then check to see if the two are equivalent:



     P | Q | P → Q | ¬Q | P ∧ ¬Q | ¬(P∧¬Q)
     T T T F F T
     T F F T T F
     F T T F F T
     F F T T F T


    So we've shown they are equivalent via truth tables.



    Now, how do we move that into the realm of natural deduction?



    The two basic options are:



    1. use inference rules

    2. do a sub proof -- either to get a contradiction or supply a conditional proof

    Since our conclusion has a negation, let's go with a subproof by contradiciton.



    1. P→Q A

    2. | P∧¬Q A

    3. | P ∧E 2

    4. | Q MP 1,3

    5. | ¬Q ∧E 2

    6. | ⊥ contradiction on 4,5

    7. ¬(P∧¬Q) RAA 2-6

    enter image description here



    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/



    (references copied from Frank Hubeny)






    share|improve this answer


























      up vote
      1
      down vote



      accepted











      ¬(P∧¬Q) from the premise P→Q




      Well, I know you stated your main issue is coming up with formal proofs (by which I'm going to assume you mean something like natural deduction) but let's just get started with the simplest conceptual level.



      1. P→Q

      this means that either P is false or Q is true based on the truth table that defines →.



      Using a truth table, we can then check to see if the two are equivalent:



       P | Q | P → Q | ¬Q | P ∧ ¬Q | ¬(P∧¬Q)
       T T T F F T
       T F F T T F
       F T T F F T
       F F T T F T


      So we've shown they are equivalent via truth tables.



      Now, how do we move that into the realm of natural deduction?



      The two basic options are:



      1. use inference rules

      2. do a sub proof -- either to get a contradiction or supply a conditional proof

      Since our conclusion has a negation, let's go with a subproof by contradiciton.



      1. P→Q A

      2. | P∧¬Q A

      3. | P ∧E 2

      4. | Q MP 1,3

      5. | ¬Q ∧E 2

      6. | ⊥ contradiction on 4,5

      7. ¬(P∧¬Q) RAA 2-6

      enter image description here



      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/



      (references copied from Frank Hubeny)






      share|improve this answer
























        up vote
        1
        down vote



        accepted







        up vote
        1
        down vote



        accepted







        ¬(P∧¬Q) from the premise P→Q




        Well, I know you stated your main issue is coming up with formal proofs (by which I'm going to assume you mean something like natural deduction) but let's just get started with the simplest conceptual level.



        1. P→Q

        this means that either P is false or Q is true based on the truth table that defines →.



        Using a truth table, we can then check to see if the two are equivalent:



         P | Q | P → Q | ¬Q | P ∧ ¬Q | ¬(P∧¬Q)
         T T T F F T
         T F F T T F
         F T T F F T
         F F T T F T


        So we've shown they are equivalent via truth tables.



        Now, how do we move that into the realm of natural deduction?



        The two basic options are:



        1. use inference rules

        2. do a sub proof -- either to get a contradiction or supply a conditional proof

        Since our conclusion has a negation, let's go with a subproof by contradiciton.



        1. P→Q A

        2. | P∧¬Q A

        3. | P ∧E 2

        4. | Q MP 1,3

        5. | ¬Q ∧E 2

        6. | ⊥ contradiction on 4,5

        7. ¬(P∧¬Q) RAA 2-6

        enter image description here



        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/



        (references copied from Frank Hubeny)






        share|improve this answer















        ¬(P∧¬Q) from the premise P→Q




        Well, I know you stated your main issue is coming up with formal proofs (by which I'm going to assume you mean something like natural deduction) but let's just get started with the simplest conceptual level.



        1. P→Q

        this means that either P is false or Q is true based on the truth table that defines →.



        Using a truth table, we can then check to see if the two are equivalent:



         P | Q | P → Q | ¬Q | P ∧ ¬Q | ¬(P∧¬Q)
         T T T F F T
         T F F T T F
         F T T F F T
         F F T T F T


        So we've shown they are equivalent via truth tables.



        Now, how do we move that into the realm of natural deduction?



        The two basic options are:



        1. use inference rules

        2. do a sub proof -- either to get a contradiction or supply a conditional proof

        Since our conclusion has a negation, let's go with a subproof by contradiciton.



        1. P→Q A

        2. | P∧¬Q A

        3. | P ∧E 2

        4. | Q MP 1,3

        5. | ¬Q ∧E 2

        6. | ⊥ contradiction on 4,5

        7. ¬(P∧¬Q) RAA 2-6

        enter image description here



        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/



        (references copied from Frank Hubeny)







        share|improve this answer














        share|improve this answer



        share|improve this answer








        edited 2 hours ago

























        answered 2 hours ago









        virmaior

        24k33793




        24k33793




















            up vote
            1
            down vote













            I agree with virmaior's proof. Since I started these as well I figured I might as well leave them here since they contain a different presentation.



            Here are two proofs using Klement's proof checker. The rules you may have to use may be different.



            enter image description here



            The proof uses conjunction elimination (∧E), conditional elimination (→E), contradiction introduction (⊥I) and negation introduction (¬I).



            To avoid the not ("¬") in front of "P∧¬Q" I started an assumption on line 2 assuming "P∧¬Q" hoping to derive a contradiction which I did on line 6. To get to that contradiction I split the conjunction into the two conjuncts, "P" and "¬Q" on lines 3 and 4. Then having "P" on line 3 and the conditional on line 1, I could put "Q" on line 5. Since lines 4 and 5 contradicted each other I could introduce a contradiction on line 6. This allowed me to discharge the assumption on line 2 and introduce the negative in from of "P∧¬Q".



            For the other one, try something like this:



            enter image description here



            The proof uses contradiction introduction (⊥I), explosion (X), disjunction elimination (∨E) and conditional introduction (→I).



            Since the goal is a conditional "¬Q → (R→P)", I assume "¬Q" on line 2 and hope to derive "R→P". Since what I want to derive is a conditional as well, I assume the premise of that conditional which is "R" on line 3. At this point I need to derive "P".



            There is a "P" on line 1, but that is a disjunction. In order to get that "P", I have to deal with the "Q" as well. The first case is easy. I have to start a subproof with just "P" as the assumption and the conclusion. This subproof is just one item, line 4. Having taken care of "P", I then have to take care of "Q". Fortunately line 2 and 5 contradict each other which allows me to introduce a contradiction on line 6. Having a contradiction I can use explosion to get anything whatsoever. I need "P", so I put "P" on line 7. This allows me to discharge both the "P" assumption on line 4 and the "Q" assumption on line 5 by eliminating the disjunction of line 1 on line 8.



            At this point note that I have a set up with "R" on line 3 and "P" on line 8. That means I can discharge the assumption of "R" on line 3 and put "R→P" on line 9. That allows me to discharge the assumption "¬Q" on line 2 and get the final result on line 10.



            You can use this proof checker online to help clarify the details yourself. The accompanying text forall x is available for further information. What you are required to use may be different. You will likely need to understand the above proofs well enough to present it in the format you need.



            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






















            • Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful!
              – Yanik
              33 mins ago














            up vote
            1
            down vote













            I agree with virmaior's proof. Since I started these as well I figured I might as well leave them here since they contain a different presentation.



            Here are two proofs using Klement's proof checker. The rules you may have to use may be different.



            enter image description here



            The proof uses conjunction elimination (∧E), conditional elimination (→E), contradiction introduction (⊥I) and negation introduction (¬I).



            To avoid the not ("¬") in front of "P∧¬Q" I started an assumption on line 2 assuming "P∧¬Q" hoping to derive a contradiction which I did on line 6. To get to that contradiction I split the conjunction into the two conjuncts, "P" and "¬Q" on lines 3 and 4. Then having "P" on line 3 and the conditional on line 1, I could put "Q" on line 5. Since lines 4 and 5 contradicted each other I could introduce a contradiction on line 6. This allowed me to discharge the assumption on line 2 and introduce the negative in from of "P∧¬Q".



            For the other one, try something like this:



            enter image description here



            The proof uses contradiction introduction (⊥I), explosion (X), disjunction elimination (∨E) and conditional introduction (→I).



            Since the goal is a conditional "¬Q → (R→P)", I assume "¬Q" on line 2 and hope to derive "R→P". Since what I want to derive is a conditional as well, I assume the premise of that conditional which is "R" on line 3. At this point I need to derive "P".



            There is a "P" on line 1, but that is a disjunction. In order to get that "P", I have to deal with the "Q" as well. The first case is easy. I have to start a subproof with just "P" as the assumption and the conclusion. This subproof is just one item, line 4. Having taken care of "P", I then have to take care of "Q". Fortunately line 2 and 5 contradict each other which allows me to introduce a contradiction on line 6. Having a contradiction I can use explosion to get anything whatsoever. I need "P", so I put "P" on line 7. This allows me to discharge both the "P" assumption on line 4 and the "Q" assumption on line 5 by eliminating the disjunction of line 1 on line 8.



            At this point note that I have a set up with "R" on line 3 and "P" on line 8. That means I can discharge the assumption of "R" on line 3 and put "R→P" on line 9. That allows me to discharge the assumption "¬Q" on line 2 and get the final result on line 10.



            You can use this proof checker online to help clarify the details yourself. The accompanying text forall x is available for further information. What you are required to use may be different. You will likely need to understand the above proofs well enough to present it in the format you need.



            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






















            • Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful!
              – Yanik
              33 mins ago












            up vote
            1
            down vote










            up vote
            1
            down vote









            I agree with virmaior's proof. Since I started these as well I figured I might as well leave them here since they contain a different presentation.



            Here are two proofs using Klement's proof checker. The rules you may have to use may be different.



            enter image description here



            The proof uses conjunction elimination (∧E), conditional elimination (→E), contradiction introduction (⊥I) and negation introduction (¬I).



            To avoid the not ("¬") in front of "P∧¬Q" I started an assumption on line 2 assuming "P∧¬Q" hoping to derive a contradiction which I did on line 6. To get to that contradiction I split the conjunction into the two conjuncts, "P" and "¬Q" on lines 3 and 4. Then having "P" on line 3 and the conditional on line 1, I could put "Q" on line 5. Since lines 4 and 5 contradicted each other I could introduce a contradiction on line 6. This allowed me to discharge the assumption on line 2 and introduce the negative in from of "P∧¬Q".



            For the other one, try something like this:



            enter image description here



            The proof uses contradiction introduction (⊥I), explosion (X), disjunction elimination (∨E) and conditional introduction (→I).



            Since the goal is a conditional "¬Q → (R→P)", I assume "¬Q" on line 2 and hope to derive "R→P". Since what I want to derive is a conditional as well, I assume the premise of that conditional which is "R" on line 3. At this point I need to derive "P".



            There is a "P" on line 1, but that is a disjunction. In order to get that "P", I have to deal with the "Q" as well. The first case is easy. I have to start a subproof with just "P" as the assumption and the conclusion. This subproof is just one item, line 4. Having taken care of "P", I then have to take care of "Q". Fortunately line 2 and 5 contradict each other which allows me to introduce a contradiction on line 6. Having a contradiction I can use explosion to get anything whatsoever. I need "P", so I put "P" on line 7. This allows me to discharge both the "P" assumption on line 4 and the "Q" assumption on line 5 by eliminating the disjunction of line 1 on line 8.



            At this point note that I have a set up with "R" on line 3 and "P" on line 8. That means I can discharge the assumption of "R" on line 3 and put "R→P" on line 9. That allows me to discharge the assumption "¬Q" on line 2 and get the final result on line 10.



            You can use this proof checker online to help clarify the details yourself. The accompanying text forall x is available for further information. What you are required to use may be different. You will likely need to understand the above proofs well enough to present it in the format you need.



            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














            I agree with virmaior's proof. Since I started these as well I figured I might as well leave them here since they contain a different presentation.



            Here are two proofs using Klement's proof checker. The rules you may have to use may be different.



            enter image description here



            The proof uses conjunction elimination (∧E), conditional elimination (→E), contradiction introduction (⊥I) and negation introduction (¬I).



            To avoid the not ("¬") in front of "P∧¬Q" I started an assumption on line 2 assuming "P∧¬Q" hoping to derive a contradiction which I did on line 6. To get to that contradiction I split the conjunction into the two conjuncts, "P" and "¬Q" on lines 3 and 4. Then having "P" on line 3 and the conditional on line 1, I could put "Q" on line 5. Since lines 4 and 5 contradicted each other I could introduce a contradiction on line 6. This allowed me to discharge the assumption on line 2 and introduce the negative in from of "P∧¬Q".



            For the other one, try something like this:



            enter image description here



            The proof uses contradiction introduction (⊥I), explosion (X), disjunction elimination (∨E) and conditional introduction (→I).



            Since the goal is a conditional "¬Q → (R→P)", I assume "¬Q" on line 2 and hope to derive "R→P". Since what I want to derive is a conditional as well, I assume the premise of that conditional which is "R" on line 3. At this point I need to derive "P".



            There is a "P" on line 1, but that is a disjunction. In order to get that "P", I have to deal with the "Q" as well. The first case is easy. I have to start a subproof with just "P" as the assumption and the conclusion. This subproof is just one item, line 4. Having taken care of "P", I then have to take care of "Q". Fortunately line 2 and 5 contradict each other which allows me to introduce a contradiction on line 6. Having a contradiction I can use explosion to get anything whatsoever. I need "P", so I put "P" on line 7. This allows me to discharge both the "P" assumption on line 4 and the "Q" assumption on line 5 by eliminating the disjunction of line 1 on line 8.



            At this point note that I have a set up with "R" on line 3 and "P" on line 8. That means I can discharge the assumption of "R" on line 3 and put "R→P" on line 9. That allows me to discharge the assumption "¬Q" on line 2 and get the final result on line 10.



            You can use this proof checker online to help clarify the details yourself. The accompanying text forall x is available for further information. What you are required to use may be different. You will likely need to understand the above proofs well enough to present it in the format you need.



            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








            edited 1 hour ago

























            answered 1 hour ago









            Frank Hubeny

            3,9943837




            3,9943837











            • Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful!
              – Yanik
              33 mins ago
















            • Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful!
              – Yanik
              33 mins ago















            Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful!
            – Yanik
            33 mins ago




            Thank you @virmaior and FrankHubeny. I'm new to this and still need a lot of practice as I see - and reading on the material conditional symbol. Your explanations were extremely helpful!
            – Yanik
            33 mins ago










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