Mixing
Logical equivalences for $P implies Q$ and $neg Q implies neg P$
$begingroup$
This is probably quite a basic question, but it's something I'm having trouble wrapping my head around.
I came across a proof which employed the following strategy
The goal was to prove
$$P implies Q$$
And the strategy used was to prove the following instead:
$$neg Q implies neg P$$
If possible, could someone provide some intuition as to why these two statements would be equivalent?
logic proof-writing proof-explanation
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
$endgroup$
|
show 2 more comments
$begingroup$
This is probably quite a basic question, but it's something I'm having trouble wrapping my head around.
I came across a proof which employed the following strategy
The goal was to prove
$$P implies Q$$
And the strategy used was to prove the following instead:
$$neg Q implies neg P$$
If possible, could someone provide some intuition as to why these two statements would be equivalent?
logic proof-writing proof-explanation
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
$endgroup$
2
$begingroup$
Why not consider some concrete examples?
$endgroup$
– Derek Elkins
Jan 31 at 23:47
2
$begingroup$
IMHO the best way to get an intuitive feeling for this kind of thing is to look at examples. "If you live in London then you live in the UK" and "if you don't live in the UK then you don't live in London" are really just two ways of stating the same fact. Make up your own examples too!
$endgroup$
– David
Jan 31 at 23:48
2
$begingroup$
This is called the contrapositive. The Wikipedia page has an intuitive explanation, though a few examples should clear this up for you.
$endgroup$
– Alexandros
Jan 31 at 23:49
$begingroup$
@David Ah okay I see, I think I understand it better if I draw a Venn Diagram - P should be contained in Q?
$endgroup$
– Sean Lee
Jan 31 at 23:51
$begingroup$
Mmm... I guess... if it works for you. Wouldn't go that way myself, doubt that it would give me any intuitive understanding.
$endgroup$
– David
Feb 1 at 0:02
|
show 2 more comments
$begingroup$
This is probably quite a basic question, but it's something I'm having trouble wrapping my head around.
I came across a proof which employed the following strategy
The goal was to prove
$$P implies Q$$
And the strategy used was to prove the following instead:
$$neg Q implies neg P$$
If possible, could someone provide some intuition as to why these two statements would be equivalent?
logic proof-writing proof-explanation
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
$endgroup$
This is probably quite a basic question, but it's something I'm having trouble wrapping my head around.
I came across a proof which employed the following strategy
The goal was to prove
$$P implies Q$$
And the strategy used was to prove the following instead:
$$neg Q implies neg P$$
If possible, could someone provide some intuition as to why these two statements would be equivalent?
logic proof-writing proof-explanation
logic proof-writing proof-explanation
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
asked Jan 31 at 23:44
Sean LeeSean Lee
749214
749214
2
$begingroup$
Why not consider some concrete examples?
$endgroup$
– Derek Elkins
Jan 31 at 23:47
2
$begingroup$
IMHO the best way to get an intuitive feeling for this kind of thing is to look at examples. "If you live in London then you live in the UK" and "if you don't live in the UK then you don't live in London" are really just two ways of stating the same fact. Make up your own examples too!
$endgroup$
– David
Jan 31 at 23:48
2
$begingroup$
This is called the contrapositive. The Wikipedia page has an intuitive explanation, though a few examples should clear this up for you.
$endgroup$
– Alexandros
Jan 31 at 23:49
$begingroup$
@David Ah okay I see, I think I understand it better if I draw a Venn Diagram - P should be contained in Q?
$endgroup$
– Sean Lee
Jan 31 at 23:51
$begingroup$
Mmm... I guess... if it works for you. Wouldn't go that way myself, doubt that it would give me any intuitive understanding.
$endgroup$
– David
Feb 1 at 0:02
|
show 2 more comments
2
$begingroup$
Why not consider some concrete examples?
$endgroup$
– Derek Elkins
Jan 31 at 23:47
2
$begingroup$
IMHO the best way to get an intuitive feeling for this kind of thing is to look at examples. "If you live in London then you live in the UK" and "if you don't live in the UK then you don't live in London" are really just two ways of stating the same fact. Make up your own examples too!
$endgroup$
– David
Jan 31 at 23:48
2
$begingroup$
This is called the contrapositive. The Wikipedia page has an intuitive explanation, though a few examples should clear this up for you.
$endgroup$
– Alexandros
Jan 31 at 23:49
$begingroup$
@David Ah okay I see, I think I understand it better if I draw a Venn Diagram - P should be contained in Q?
$endgroup$
– Sean Lee
Jan 31 at 23:51
$begingroup$
Mmm... I guess... if it works for you. Wouldn't go that way myself, doubt that it would give me any intuitive understanding.
$endgroup$
– David
Feb 1 at 0:02
2
2
$begingroup$
Why not consider some concrete examples?
$endgroup$
– Derek Elkins
Jan 31 at 23:47
$begingroup$
Why not consider some concrete examples?
$endgroup$
– Derek Elkins
Jan 31 at 23:47
2
2
$begingroup$
IMHO the best way to get an intuitive feeling for this kind of thing is to look at examples. "If you live in London then you live in the UK" and "if you don't live in the UK then you don't live in London" are really just two ways of stating the same fact. Make up your own examples too!
$endgroup$
– David
Jan 31 at 23:48
$begingroup$
IMHO the best way to get an intuitive feeling for this kind of thing is to look at examples. "If you live in London then you live in the UK" and "if you don't live in the UK then you don't live in London" are really just two ways of stating the same fact. Make up your own examples too!
$endgroup$
– David
Jan 31 at 23:48
2
2
$begingroup$
This is called the contrapositive. The Wikipedia page has an intuitive explanation, though a few examples should clear this up for you.
$endgroup$
– Alexandros
Jan 31 at 23:49
$begingroup$
This is called the contrapositive. The Wikipedia page has an intuitive explanation, though a few examples should clear this up for you.
$endgroup$
– Alexandros
Jan 31 at 23:49
$begingroup$
@David Ah okay I see, I think I understand it better if I draw a Venn Diagram - P should be contained in Q?
$endgroup$
– Sean Lee
Jan 31 at 23:51
$begingroup$
@David Ah okay I see, I think I understand it better if I draw a Venn Diagram - P should be contained in Q?
$endgroup$
– Sean Lee
Jan 31 at 23:51
$begingroup$
Mmm... I guess... if it works for you. Wouldn't go that way myself, doubt that it would give me any intuitive understanding.
$endgroup$
– David
Feb 1 at 0:02
$begingroup$
Mmm... I guess... if it works for you. Wouldn't go that way myself, doubt that it would give me any intuitive understanding.
$endgroup$
– David
Feb 1 at 0:02
|
show 2 more comments
3 Answers
3
active
oldest
votes
$begingroup$
$P implies Q$ means whenever you have $P$ you will have $Q$.
That means if you have $P$ then $lnot Q$ is impossible.
That means if you do have $lnot Q$, it isn't possible that you had $P$ (because then you would have had $Q$).
That means if you have $lnot Q$ then you will have $lnot P$.
That means $lnot Q implies lnot P$.
... So if "$P implies Q$" is true it is also true that "$lnot Q implies lnot P$".
====
Likewise $lnot Q implies lnot P$ means whenever $Q$ is false then $P$ is false.
So if you have $P$ is true it isn't possible that you had $Q$ is false (because if $Q$ is false $P$ wouldn't be true.)
So if somehow you had $P$ is true and it isn't possible that $Q$ is false, it must be that $Q$ is true.
So if you did have $P$ is true it must follow that $Q$ is true.
So $P implies Q$.
... So if "$lnot Q implies lnot P$" is true it will follow that "$P implies Q$" is true.
====
So the two statements $Pimplies Q$ and $lnot Q implies lnot P$ can only be true if the other one is true, and if one is false the other can't be true and if one is true the other must also be true, these two statements are always true or false under the exact same circumstances.
In other words... they are equivalent statements.
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
$endgroup$
$begingroup$
Wow, this really cleared things up for me! Thank you for going through it using first principles, it becomes so much clearer! (:
$endgroup$
– Sean Lee
Feb 1 at 0:18
add a comment |
$begingroup$
In classical logic this boils down to the definition of $implies$. You can then check this equivalence by testing all possible combinations of truth values for $P$ and $Q$. This principle is called contraposition.
In intuitionistic logic, you only get that $Pimplies Q$ implies $lnot Qimplies lnot P$, but not the converse.
Contraposition is often a convenient way of starting a proof, in particular if you don't know where you want to go with your proof. But many such proofs can be in fact reformulated to be direct proofs.
answered Jan 31 at 23:58
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
If $P$ implies $Q$, then $lnot Q$ cannot imply $P$; because that would in turn imply $Q$ (a contradiction). So therefore must $lnot Q$ imply $lnot P$.
That is: $Pto Q$ entails that $lnot Qtolnot P$.
Likewise if $lnot Q$ implies $lnot P$, then $P$ cannot imply $lnot Q$; because that would in turn imply $lnot P$ (a contradiction). So therefore must $P$ imply $lnotlnot Q$.
In classical logic, that double negation can be eliminated, so we can say $lnot Qtolnot P$ entails $Pto Q$.
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
add a comment |
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3 Answers
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3 Answers
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$begingroup$
$P implies Q$ means whenever you have $P$ you will have $Q$.
That means if you have $P$ then $lnot Q$ is impossible.
That means if you do have $lnot Q$, it isn't possible that you had $P$ (because then you would have had $Q$).
That means if you have $lnot Q$ then you will have $lnot P$.
That means $lnot Q implies lnot P$.
... So if "$P implies Q$" is true it is also true that "$lnot Q implies lnot P$".
====
Likewise $lnot Q implies lnot P$ means whenever $Q$ is false then $P$ is false.
So if you have $P$ is true it isn't possible that you had $Q$ is false (because if $Q$ is false $P$ wouldn't be true.)
So if somehow you had $P$ is true and it isn't possible that $Q$ is false, it must be that $Q$ is true.
So if you did have $P$ is true it must follow that $Q$ is true.
So $P implies Q$.
... So if "$lnot Q implies lnot P$" is true it will follow that "$P implies Q$" is true.
====
So the two statements $Pimplies Q$ and $lnot Q implies lnot P$ can only be true if the other one is true, and if one is false the other can't be true and if one is true the other must also be true, these two statements are always true or false under the exact same circumstances.
In other words... they are equivalent statements.
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
$endgroup$
$begingroup$
Wow, this really cleared things up for me! Thank you for going through it using first principles, it becomes so much clearer! (:
$endgroup$
– Sean Lee
Feb 1 at 0:18
add a comment |
$begingroup$
$P implies Q$ means whenever you have $P$ you will have $Q$.
That means if you have $P$ then $lnot Q$ is impossible.
That means if you do have $lnot Q$, it isn't possible that you had $P$ (because then you would have had $Q$).
That means if you have $lnot Q$ then you will have $lnot P$.
That means $lnot Q implies lnot P$.
... So if "$P implies Q$" is true it is also true that "$lnot Q implies lnot P$".
====
Likewise $lnot Q implies lnot P$ means whenever $Q$ is false then $P$ is false.
So if you have $P$ is true it isn't possible that you had $Q$ is false (because if $Q$ is false $P$ wouldn't be true.)
So if somehow you had $P$ is true and it isn't possible that $Q$ is false, it must be that $Q$ is true.
So if you did have $P$ is true it must follow that $Q$ is true.
So $P implies Q$.
... So if "$lnot Q implies lnot P$" is true it will follow that "$P implies Q$" is true.
====
So the two statements $Pimplies Q$ and $lnot Q implies lnot P$ can only be true if the other one is true, and if one is false the other can't be true and if one is true the other must also be true, these two statements are always true or false under the exact same circumstances.
In other words... they are equivalent statements.
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
$endgroup$
$begingroup$
Wow, this really cleared things up for me! Thank you for going through it using first principles, it becomes so much clearer! (:
$endgroup$
– Sean Lee
Feb 1 at 0:18
add a comment |
$begingroup$
$P implies Q$ means whenever you have $P$ you will have $Q$.
That means if you have $P$ then $lnot Q$ is impossible.
That means if you do have $lnot Q$, it isn't possible that you had $P$ (because then you would have had $Q$).
That means if you have $lnot Q$ then you will have $lnot P$.
That means $lnot Q implies lnot P$.
... So if "$P implies Q$" is true it is also true that "$lnot Q implies lnot P$".
====
Likewise $lnot Q implies lnot P$ means whenever $Q$ is false then $P$ is false.
So if you have $P$ is true it isn't possible that you had $Q$ is false (because if $Q$ is false $P$ wouldn't be true.)
So if somehow you had $P$ is true and it isn't possible that $Q$ is false, it must be that $Q$ is true.
So if you did have $P$ is true it must follow that $Q$ is true.
So $P implies Q$.
... So if "$lnot Q implies lnot P$" is true it will follow that "$P implies Q$" is true.
====
So the two statements $Pimplies Q$ and $lnot Q implies lnot P$ can only be true if the other one is true, and if one is false the other can't be true and if one is true the other must also be true, these two statements are always true or false under the exact same circumstances.
In other words... they are equivalent statements.
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
$endgroup$
$P implies Q$ means whenever you have $P$ you will have $Q$.
That means if you have $P$ then $lnot Q$ is impossible.
That means if you do have $lnot Q$, it isn't possible that you had $P$ (because then you would have had $Q$).
That means if you have $lnot Q$ then you will have $lnot P$.
That means $lnot Q implies lnot P$.
... So if "$P implies Q$" is true it is also true that "$lnot Q implies lnot P$".
====
Likewise $lnot Q implies lnot P$ means whenever $Q$ is false then $P$ is false.
So if you have $P$ is true it isn't possible that you had $Q$ is false (because if $Q$ is false $P$ wouldn't be true.)
So if somehow you had $P$ is true and it isn't possible that $Q$ is false, it must be that $Q$ is true.
So if you did have $P$ is true it must follow that $Q$ is true.
So $P implies Q$.
... So if "$lnot Q implies lnot P$" is true it will follow that "$P implies Q$" is true.
====
So the two statements $Pimplies Q$ and $lnot Q implies lnot P$ can only be true if the other one is true, and if one is false the other can't be true and if one is true the other must also be true, these two statements are always true or false under the exact same circumstances.
In other words... they are equivalent statements.
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
answered Feb 1 at 0:16
fleabloodfleablood
73.9k22891
73.9k22891
$begingroup$
Wow, this really cleared things up for me! Thank you for going through it using first principles, it becomes so much clearer! (:
$endgroup$
– Sean Lee
Feb 1 at 0:18
add a comment |
$begingroup$
Wow, this really cleared things up for me! Thank you for going through it using first principles, it becomes so much clearer! (:
$endgroup$
– Sean Lee
Feb 1 at 0:18
$begingroup$
Wow, this really cleared things up for me! Thank you for going through it using first principles, it becomes so much clearer! (:
$endgroup$
– Sean Lee
Feb 1 at 0:18
$begingroup$
Wow, this really cleared things up for me! Thank you for going through it using first principles, it becomes so much clearer! (:
$endgroup$
– Sean Lee
Feb 1 at 0:18
add a comment |
$begingroup$
In classical logic this boils down to the definition of $implies$. You can then check this equivalence by testing all possible combinations of truth values for $P$ and $Q$. This principle is called contraposition.
In intuitionistic logic, you only get that $Pimplies Q$ implies $lnot Qimplies lnot P$, but not the converse.
Contraposition is often a convenient way of starting a proof, in particular if you don't know where you want to go with your proof. But many such proofs can be in fact reformulated to be direct proofs.
answered Jan 31 at 23:58
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
In classical logic this boils down to the definition of $implies$. You can then check this equivalence by testing all possible combinations of truth values for $P$ and $Q$. This principle is called contraposition.
In intuitionistic logic, you only get that $Pimplies Q$ implies $lnot Qimplies lnot P$, but not the converse.
Contraposition is often a convenient way of starting a proof, in particular if you don't know where you want to go with your proof. But many such proofs can be in fact reformulated to be direct proofs.
answered Jan 31 at 23:58
ffffforallffffforall
36028
$endgroup$
add a comment |
$begingroup$
In classical logic this boils down to the definition of $implies$. You can then check this equivalence by testing all possible combinations of truth values for $P$ and $Q$. This principle is called contraposition.
In intuitionistic logic, you only get that $Pimplies Q$ implies $lnot Qimplies lnot P$, but not the converse.
Contraposition is often a convenient way of starting a proof, in particular if you don't know where you want to go with your proof. But many such proofs can be in fact reformulated to be direct proofs.
answered Jan 31 at 23:58
ffffforallffffforall
36028
$endgroup$
In classical logic this boils down to the definition of $implies$. You can then check this equivalence by testing all possible combinations of truth values for $P$ and $Q$. This principle is called contraposition.
In intuitionistic logic, you only get that $Pimplies Q$ implies $lnot Qimplies lnot P$, but not the converse.
Contraposition is often a convenient way of starting a proof, in particular if you don't know where you want to go with your proof. But many such proofs can be in fact reformulated to be direct proofs.
answered Jan 31 at 23:58
ffffforallffffforall
36028
answered Jan 31 at 23:58
ffffforallffffforall
36028
answered Jan 31 at 23:58
ffffforallffffforall
36028
answered Jan 31 at 23:58
ffffforallffffforall
36028
36028
add a comment |
add a comment |
$begingroup$
If $P$ implies $Q$, then $lnot Q$ cannot imply $P$; because that would in turn imply $Q$ (a contradiction). So therefore must $lnot Q$ imply $lnot P$.
That is: $Pto Q$ entails that $lnot Qtolnot P$.
Likewise if $lnot Q$ implies $lnot P$, then $P$ cannot imply $lnot Q$; because that would in turn imply $lnot P$ (a contradiction). So therefore must $P$ imply $lnotlnot Q$.
In classical logic, that double negation can be eliminated, so we can say $lnot Qtolnot P$ entails $Pto Q$.
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
add a comment |
$begingroup$
If $P$ implies $Q$, then $lnot Q$ cannot imply $P$; because that would in turn imply $Q$ (a contradiction). So therefore must $lnot Q$ imply $lnot P$.
That is: $Pto Q$ entails that $lnot Qtolnot P$.
Likewise if $lnot Q$ implies $lnot P$, then $P$ cannot imply $lnot Q$; because that would in turn imply $lnot P$ (a contradiction). So therefore must $P$ imply $lnotlnot Q$.
In classical logic, that double negation can be eliminated, so we can say $lnot Qtolnot P$ entails $Pto Q$.
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
add a comment |
$begingroup$
If $P$ implies $Q$, then $lnot Q$ cannot imply $P$; because that would in turn imply $Q$ (a contradiction). So therefore must $lnot Q$ imply $lnot P$.
That is: $Pto Q$ entails that $lnot Qtolnot P$.
Likewise if $lnot Q$ implies $lnot P$, then $P$ cannot imply $lnot Q$; because that would in turn imply $lnot P$ (a contradiction). So therefore must $P$ imply $lnotlnot Q$.
In classical logic, that double negation can be eliminated, so we can say $lnot Qtolnot P$ entails $Pto Q$.
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
$endgroup$
If $P$ implies $Q$, then $lnot Q$ cannot imply $P$; because that would in turn imply $Q$ (a contradiction). So therefore must $lnot Q$ imply $lnot P$.
That is: $Pto Q$ entails that $lnot Qtolnot P$.
Likewise if $lnot Q$ implies $lnot P$, then $P$ cannot imply $lnot Q$; because that would in turn imply $lnot P$ (a contradiction). So therefore must $P$ imply $lnotlnot Q$.
In classical logic, that double negation can be eliminated, so we can say $lnot Qtolnot P$ entails $Pto Q$.
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
answered Feb 1 at 0:18
Graham KempGraham Kemp
87.8k43578
87.8k43578
add a comment |
add a comment |
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2
$begingroup$
Why not consider some concrete examples?
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– Derek Elkins
Jan 31 at 23:47
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IMHO the best way to get an intuitive feeling for this kind of thing is to look at examples. "If you live in London then you live in the UK" and "if you don't live in the UK then you don't live in London" are really just two ways of stating the same fact. Make up your own examples too!
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– David
Jan 31 at 23:48
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This is called the contrapositive. The Wikipedia page has an intuitive explanation, though a few examples should clear this up for you.
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– Alexandros
Jan 31 at 23:49
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@David Ah okay I see, I think I understand it better if I draw a Venn Diagram - P should be contained in Q?
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– Sean Lee
Jan 31 at 23:51
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Mmm... I guess... if it works for you. Wouldn't go that way myself, doubt that it would give me any intuitive understanding.
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– David
Feb 1 at 0:02