Logic: Is Associativity Rule applicable in this statement (~P ^ Q) ^ P? [closed]

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Logic: Is Associativity Rule applicable in this statement (~P ^ Q) ^ P such that it will become ~P ^ (Q ^ P)?
I'm trying to prove that the statement is Contradictory and so I was thinking of doing the following as a proof:



  • (~P ^ Q) ^ P = Premise

  • ~P ^ (Q ^ P) = Associativity by Conjunction

  • ~P ^ (P) = Simplification

  • P ^ ~P = Commutativity

  • P ^ ~P = Contradictory

I'm hoping to hear your thoughts on this.
Thank you!










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closed as off-topic by Gerry Myerson, Leucippus, Shailesh, user91500, amWhy Sep 16 at 0:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Shailesh, user91500, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You mean $(neg P wedge Q)wedge P$? Then indeed you're right.
    – Wuestenfux
    Sep 10 at 9:06











  • Does it mean that it is possible to translate (¬P∧Q)∧P to ¬P∧(Q∧P) by way of Associativity Rule?
    – Jeryl Donato Estopace
    Sep 10 at 9:23










  • Indeed, the rule of associativity is applicable.
    – Wuestenfux
    Sep 10 at 9:24










  • Thank you very much!
    – Jeryl Donato Estopace
    Sep 10 at 9:27






  • 1




    But I don't know what's going on in the line you label "Simplification".
    – Gerry Myerson
    Sep 10 at 10:17














up vote
0
down vote

favorite












Logic: Is Associativity Rule applicable in this statement (~P ^ Q) ^ P such that it will become ~P ^ (Q ^ P)?
I'm trying to prove that the statement is Contradictory and so I was thinking of doing the following as a proof:



  • (~P ^ Q) ^ P = Premise

  • ~P ^ (Q ^ P) = Associativity by Conjunction

  • ~P ^ (P) = Simplification

  • P ^ ~P = Commutativity

  • P ^ ~P = Contradictory

I'm hoping to hear your thoughts on this.
Thank you!










share|cite|improve this question















closed as off-topic by Gerry Myerson, Leucippus, Shailesh, user91500, amWhy Sep 16 at 0:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Shailesh, user91500, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.












  • You mean $(neg P wedge Q)wedge P$? Then indeed you're right.
    – Wuestenfux
    Sep 10 at 9:06











  • Does it mean that it is possible to translate (¬P∧Q)∧P to ¬P∧(Q∧P) by way of Associativity Rule?
    – Jeryl Donato Estopace
    Sep 10 at 9:23










  • Indeed, the rule of associativity is applicable.
    – Wuestenfux
    Sep 10 at 9:24










  • Thank you very much!
    – Jeryl Donato Estopace
    Sep 10 at 9:27






  • 1




    But I don't know what's going on in the line you label "Simplification".
    – Gerry Myerson
    Sep 10 at 10:17












up vote
0
down vote

favorite









up vote
0
down vote

favorite











Logic: Is Associativity Rule applicable in this statement (~P ^ Q) ^ P such that it will become ~P ^ (Q ^ P)?
I'm trying to prove that the statement is Contradictory and so I was thinking of doing the following as a proof:



  • (~P ^ Q) ^ P = Premise

  • ~P ^ (Q ^ P) = Associativity by Conjunction

  • ~P ^ (P) = Simplification

  • P ^ ~P = Commutativity

  • P ^ ~P = Contradictory

I'm hoping to hear your thoughts on this.
Thank you!










share|cite|improve this question















Logic: Is Associativity Rule applicable in this statement (~P ^ Q) ^ P such that it will become ~P ^ (Q ^ P)?
I'm trying to prove that the statement is Contradictory and so I was thinking of doing the following as a proof:



  • (~P ^ Q) ^ P = Premise

  • ~P ^ (Q ^ P) = Associativity by Conjunction

  • ~P ^ (P) = Simplification

  • P ^ ~P = Commutativity

  • P ^ ~P = Contradictory

I'm hoping to hear your thoughts on this.
Thank you!







logic propositional-calculus






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edited Sep 10 at 9:25

























asked Sep 10 at 9:05









Jeryl Donato Estopace

42




42




closed as off-topic by Gerry Myerson, Leucippus, Shailesh, user91500, amWhy Sep 16 at 0:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Shailesh, user91500, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.




closed as off-topic by Gerry Myerson, Leucippus, Shailesh, user91500, amWhy Sep 16 at 0:11


This question appears to be off-topic. The users who voted to close gave this specific reason:


  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Leucippus, Shailesh, user91500, amWhy
If this question can be reworded to fit the rules in the help center, please edit the question.











  • You mean $(neg P wedge Q)wedge P$? Then indeed you're right.
    – Wuestenfux
    Sep 10 at 9:06











  • Does it mean that it is possible to translate (¬P∧Q)∧P to ¬P∧(Q∧P) by way of Associativity Rule?
    – Jeryl Donato Estopace
    Sep 10 at 9:23










  • Indeed, the rule of associativity is applicable.
    – Wuestenfux
    Sep 10 at 9:24










  • Thank you very much!
    – Jeryl Donato Estopace
    Sep 10 at 9:27






  • 1




    But I don't know what's going on in the line you label "Simplification".
    – Gerry Myerson
    Sep 10 at 10:17
















  • You mean $(neg P wedge Q)wedge P$? Then indeed you're right.
    – Wuestenfux
    Sep 10 at 9:06











  • Does it mean that it is possible to translate (¬P∧Q)∧P to ¬P∧(Q∧P) by way of Associativity Rule?
    – Jeryl Donato Estopace
    Sep 10 at 9:23










  • Indeed, the rule of associativity is applicable.
    – Wuestenfux
    Sep 10 at 9:24










  • Thank you very much!
    – Jeryl Donato Estopace
    Sep 10 at 9:27






  • 1




    But I don't know what's going on in the line you label "Simplification".
    – Gerry Myerson
    Sep 10 at 10:17















You mean $(neg P wedge Q)wedge P$? Then indeed you're right.
– Wuestenfux
Sep 10 at 9:06





You mean $(neg P wedge Q)wedge P$? Then indeed you're right.
– Wuestenfux
Sep 10 at 9:06













Does it mean that it is possible to translate (¬P∧Q)∧P to ¬P∧(Q∧P) by way of Associativity Rule?
– Jeryl Donato Estopace
Sep 10 at 9:23




Does it mean that it is possible to translate (¬P∧Q)∧P to ¬P∧(Q∧P) by way of Associativity Rule?
– Jeryl Donato Estopace
Sep 10 at 9:23












Indeed, the rule of associativity is applicable.
– Wuestenfux
Sep 10 at 9:24




Indeed, the rule of associativity is applicable.
– Wuestenfux
Sep 10 at 9:24












Thank you very much!
– Jeryl Donato Estopace
Sep 10 at 9:27




Thank you very much!
– Jeryl Donato Estopace
Sep 10 at 9:27




1




1




But I don't know what's going on in the line you label "Simplification".
– Gerry Myerson
Sep 10 at 10:17




But I don't know what's going on in the line you label "Simplification".
– Gerry Myerson
Sep 10 at 10:17















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