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!
logic propositional-calculus
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
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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!
logic propositional-calculus
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
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
 |Â
show 2 more comments
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!
logic propositional-calculus
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
logic propositional-calculus
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
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
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
 |Â
show 2 more comments
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
 |Â
show 2 more comments
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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