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Logic p and q implies p tautology [closed]
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we recently got the statement that $(Q to P) land P$ is a tautology, but i do not get why it is one, since it is wrong for certain cases of P and Q.
logic
edited Aug 6 at 0:40
Graham Kemp
80.1k43275
asked Aug 6 at 0:31
MaxMath
6
closed as unclear what you're asking by Shaun, Noah Schweber, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
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up vote
-1
down vote
favorite
we recently got the statement that $(Q to P) land P$ is a tautology, but i do not get why it is one, since it is wrong for certain cases of P and Q.
logic
edited Aug 6 at 0:40
Graham Kemp
80.1k43275
asked Aug 6 at 0:31
MaxMath
6
closed as unclear what you're asking by Shaun, Noah Schweber, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Your statement in the question is different in to the one in the title. Also, you need to add dollar signs on either side of the statement to get the MathJax working.
– Chessanator
Aug 6 at 0:40
You should double-check your notes to make sure you've read the claim correctly.
– Noah Schweber
Aug 6 at 1:00
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
we recently got the statement that $(Q to P) land P$ is a tautology, but i do not get why it is one, since it is wrong for certain cases of P and Q.
logic
edited Aug 6 at 0:40
Graham Kemp
80.1k43275
asked Aug 6 at 0:31
MaxMath
6
we recently got the statement that $(Q to P) land P$ is a tautology, but i do not get why it is one, since it is wrong for certain cases of P and Q.
logic
edited Aug 6 at 0:40
Graham Kemp
80.1k43275
asked Aug 6 at 0:31
MaxMath
6
edited Aug 6 at 0:40
Graham Kemp
80.1k43275
edited Aug 6 at 0:40
Graham Kemp
80.1k43275
edited Aug 6 at 0:40
Graham Kemp
80.1k43275
80.1k43275
asked Aug 6 at 0:31
MaxMath
6
asked Aug 6 at 0:31
MaxMath
6
asked Aug 6 at 0:31
MaxMath
6
6
closed as unclear what you're asking by Shaun, Noah Schweber, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as unclear what you're asking by Shaun, Noah Schweber, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:10
Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
2
Your statement in the question is different in to the one in the title. Also, you need to add dollar signs on either side of the statement to get the MathJax working.
– Chessanator
Aug 6 at 0:40
You should double-check your notes to make sure you've read the claim correctly.
– Noah Schweber
Aug 6 at 1:00
add a comment |Â
2
Your statement in the question is different in to the one in the title. Also, you need to add dollar signs on either side of the statement to get the MathJax working.
– Chessanator
Aug 6 at 0:40
You should double-check your notes to make sure you've read the claim correctly.
– Noah Schweber
Aug 6 at 1:00
2
2
Your statement in the question is different in to the one in the title. Also, you need to add dollar signs on either side of the statement to get the MathJax working.
– Chessanator
Aug 6 at 0:40
Your statement in the question is different in to the one in the title. Also, you need to add dollar signs on either side of the statement to get the MathJax working.
– Chessanator
Aug 6 at 0:40
You should double-check your notes to make sure you've read the claim correctly.
– Noah Schweber
Aug 6 at 1:00
You should double-check your notes to make sure you've read the claim correctly.
– Noah Schweber
Aug 6 at 1:00
add a comment |Â
1 Answer
1
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1
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$Pland Q~to~P$, also written as $(Pland Q)to P$, is a tautology. Â "If $P$ and $Q$ are both true, then $P$ will be true," is clearly so.
Should either $P$ or $Q$ be false, the implication still holds, because implications are considerer true when their antecedants are false (or their consequents are true).
An implication is only considered false when the consequent can be false while the antecedant is true. Â However, $P$ can not be false when both $P$ and $Q$ are true.
Thus for all possible values of $P$ and $Q$ the implication, $(Pland Q)to P$, will be satisfied. Â It is a tautology.
$Pland (Qto P)$ is a completely different matter. Â It is not the same thing at all, and not a tautology. Â Although it can be satisfied, it is clearly false whenever $P$ is false. Â It is a contingent statement.
Long story short: The placement of the brackets matters. Â The rules of operation precedence means "$P$ and $Q$ implies $P$" should be read as $(Pland Q)to P$.
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
$Pland Q~to~P$, also written as $(Pland Q)to P$, is a tautology. Â "If $P$ and $Q$ are both true, then $P$ will be true," is clearly so.
Should either $P$ or $Q$ be false, the implication still holds, because implications are considerer true when their antecedants are false (or their consequents are true).
An implication is only considered false when the consequent can be false while the antecedant is true. Â However, $P$ can not be false when both $P$ and $Q$ are true.
Thus for all possible values of $P$ and $Q$ the implication, $(Pland Q)to P$, will be satisfied. Â It is a tautology.
$Pland (Qto P)$ is a completely different matter. Â It is not the same thing at all, and not a tautology. Â Although it can be satisfied, it is clearly false whenever $P$ is false. Â It is a contingent statement.
Long story short: The placement of the brackets matters. Â The rules of operation precedence means "$P$ and $Q$ implies $P$" should be read as $(Pland Q)to P$.
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
add a comment |Â
up vote
1
down vote
$Pland Q~to~P$, also written as $(Pland Q)to P$, is a tautology. Â "If $P$ and $Q$ are both true, then $P$ will be true," is clearly so.
Should either $P$ or $Q$ be false, the implication still holds, because implications are considerer true when their antecedants are false (or their consequents are true).
An implication is only considered false when the consequent can be false while the antecedant is true. Â However, $P$ can not be false when both $P$ and $Q$ are true.
Thus for all possible values of $P$ and $Q$ the implication, $(Pland Q)to P$, will be satisfied. Â It is a tautology.
$Pland (Qto P)$ is a completely different matter. Â It is not the same thing at all, and not a tautology. Â Although it can be satisfied, it is clearly false whenever $P$ is false. Â It is a contingent statement.
Long story short: The placement of the brackets matters. Â The rules of operation precedence means "$P$ and $Q$ implies $P$" should be read as $(Pland Q)to P$.
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
add a comment |Â
up vote
1
down vote
up vote
1
down vote
$Pland Q~to~P$, also written as $(Pland Q)to P$, is a tautology. Â "If $P$ and $Q$ are both true, then $P$ will be true," is clearly so.
Should either $P$ or $Q$ be false, the implication still holds, because implications are considerer true when their antecedants are false (or their consequents are true).
An implication is only considered false when the consequent can be false while the antecedant is true. Â However, $P$ can not be false when both $P$ and $Q$ are true.
Thus for all possible values of $P$ and $Q$ the implication, $(Pland Q)to P$, will be satisfied. Â It is a tautology.
$Pland (Qto P)$ is a completely different matter. Â It is not the same thing at all, and not a tautology. Â Although it can be satisfied, it is clearly false whenever $P$ is false. Â It is a contingent statement.
Long story short: The placement of the brackets matters. Â The rules of operation precedence means "$P$ and $Q$ implies $P$" should be read as $(Pland Q)to P$.
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
$Pland Q~to~P$, also written as $(Pland Q)to P$, is a tautology. Â "If $P$ and $Q$ are both true, then $P$ will be true," is clearly so.
Should either $P$ or $Q$ be false, the implication still holds, because implications are considerer true when their antecedants are false (or their consequents are true).
An implication is only considered false when the consequent can be false while the antecedant is true. Â However, $P$ can not be false when both $P$ and $Q$ are true.
Thus for all possible values of $P$ and $Q$ the implication, $(Pland Q)to P$, will be satisfied. Â It is a tautology.
$Pland (Qto P)$ is a completely different matter. Â It is not the same thing at all, and not a tautology. Â Although it can be satisfied, it is clearly false whenever $P$ is false. Â It is a contingent statement.
Long story short: The placement of the brackets matters. Â The rules of operation precedence means "$P$ and $Q$ implies $P$" should be read as $(Pland Q)to P$.
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
edited Aug 6 at 0:59
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
answered Aug 6 at 0:53
Graham Kemp
80.1k43275
80.1k43275
add a comment |Â
add a comment |Â
2
Your statement in the question is different in to the one in the title. Also, you need to add dollar signs on either side of the statement to get the MathJax working.
– Chessanator
Aug 6 at 0:40
You should double-check your notes to make sure you've read the claim correctly.
– Noah Schweber
Aug 6 at 1:00