Confused about proofs by contradiction, the Law of the Excluded Middle and existence of consistent axiomatic systems.True vs. ProvableWhy not both true and false?Why doesn't the independence of the continuum hypothesis immediately imply that ZFC is unsatisfactory?Why is establishing absolute consistency of ZFC impossible?How could a statement be true without proof?Provability of the continuum hypothesis and the incompleteness theoremWhat is the meaning of “true”?Absoluteness of $ textCon(mathsfZFC) $ for Transitive Models of $ mathsfZFC $.Contraposition and law of excluded middleHow is the law of excluded middle necessary for proofs by contradiction?Is there any case where classical logic has “proven” an incorrect result?Systems without the law of excluded middleWill assuming an undecidable statement result in a consistent system?Trying to understand self-reference as it relates to Godel's Second Incompleteness TheoremShowing propositional logic is consistentHow can a formal system ever be non-obviously unsound?Propositional calculus and intuitionist logic
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Confused about proofs by contradiction, the Law of the Excluded Middle and existence of consistent axiomatic systems.
True vs. ProvableWhy not both true and false?Why doesn't the independence of the continuum hypothesis immediately imply that ZFC is unsatisfactory?Why is establishing absolute consistency of ZFC impossible?How could a statement be true without proof?Provability of the continuum hypothesis and the incompleteness theoremWhat is the meaning of “true”?Absoluteness of $ textCon(mathsfZFC) $ for Transitive Models of $ mathsfZFC $.Contraposition and law of excluded middleHow is the law of excluded middle necessary for proofs by contradiction?Is there any case where classical logic has “proven” an incorrect result?Systems without the law of excluded middleWill assuming an undecidable statement result in a consistent system?Trying to understand self-reference as it relates to Godel's Second Incompleteness TheoremShowing propositional logic is consistentHow can a formal system ever be non-obviously unsound?Propositional calculus and intuitionist logic
$begingroup$
I apologize if my question is too dumb. I'm not particularly educated in this area of Mathematics.
Proof by contradiction consists of assuming a statement $P$ is false, and then reach a contradiction thus allowing us to conclude that $P$ must be true. Such line of reasoning seems to be using the Law of the Excluded Middle, that is, $P lor neg P$ is a tautology.
Wouldn't assuming said law lead to some problems. As an example, it has been proven that if ZFC is consistent, then both ZFC$+$CH and ZFC$+neg$CH are also consistent. Thus, by LEM, there are only two possible options:
1) CH is true, but unprovable within ZFC.
2) $neg$CH is true, but unprovable within ZFC.
Suppose for a second that the first option was correct. Since $neg$ CH is consistent with ZFC, the axiomatic system ZFC$+ neg$CH contains no contradictions. However CH being true does imply that ZFC$+ neg$CH has a contradiction. The second option being true leads to the same result.
What am I missing?
I would truly appreciate any help/thoughts.
logic soft-question
$endgroup$
|
show 4 more comments
$begingroup$
I apologize if my question is too dumb. I'm not particularly educated in this area of Mathematics.
Proof by contradiction consists of assuming a statement $P$ is false, and then reach a contradiction thus allowing us to conclude that $P$ must be true. Such line of reasoning seems to be using the Law of the Excluded Middle, that is, $P lor neg P$ is a tautology.
Wouldn't assuming said law lead to some problems. As an example, it has been proven that if ZFC is consistent, then both ZFC$+$CH and ZFC$+neg$CH are also consistent. Thus, by LEM, there are only two possible options:
1) CH is true, but unprovable within ZFC.
2) $neg$CH is true, but unprovable within ZFC.
Suppose for a second that the first option was correct. Since $neg$ CH is consistent with ZFC, the axiomatic system ZFC$+ neg$CH contains no contradictions. However CH being true does imply that ZFC$+ neg$CH has a contradiction. The second option being true leads to the same result.
What am I missing?
I would truly appreciate any help/thoughts.
logic soft-question
$endgroup$
1
$begingroup$
It doesn't make sense to say CH is true in ZFC. CH is true in some models of ZFC, and it's false in other models of ZFC. (Assuming there are any models of ZFC...)
$endgroup$
– Derek Elkins
Apr 30 at 20:45
$begingroup$
what do you exactly mean by "true in ZFC"?
$endgroup$
– Zuhair
Apr 30 at 20:45
$begingroup$
You have in the title a very broad subject, and the body of your Question narrows this only in a fairly confused way. How would you define "CH is true in ZFC"? It might be more productive to begin your investigations on grounds where you have a clear understanding, rather than by combining several topics that are muddled.
$endgroup$
– hardmath
Apr 30 at 20:45
$begingroup$
I meant to say that 'CH is true along the axioms of ZFC'. I'll edit my question. I'm sorry, I'm not familiar with the correct terminology.
$endgroup$
– Leo
Apr 30 at 20:51
$begingroup$
You may be interesting in reading this question and an excellent response: math.stackexchange.com/questions/1746563/…
$endgroup$
– Vasya
Apr 30 at 20:51
|
show 4 more comments
$begingroup$
I apologize if my question is too dumb. I'm not particularly educated in this area of Mathematics.
Proof by contradiction consists of assuming a statement $P$ is false, and then reach a contradiction thus allowing us to conclude that $P$ must be true. Such line of reasoning seems to be using the Law of the Excluded Middle, that is, $P lor neg P$ is a tautology.
Wouldn't assuming said law lead to some problems. As an example, it has been proven that if ZFC is consistent, then both ZFC$+$CH and ZFC$+neg$CH are also consistent. Thus, by LEM, there are only two possible options:
1) CH is true, but unprovable within ZFC.
2) $neg$CH is true, but unprovable within ZFC.
Suppose for a second that the first option was correct. Since $neg$ CH is consistent with ZFC, the axiomatic system ZFC$+ neg$CH contains no contradictions. However CH being true does imply that ZFC$+ neg$CH has a contradiction. The second option being true leads to the same result.
What am I missing?
I would truly appreciate any help/thoughts.
logic soft-question
$endgroup$
I apologize if my question is too dumb. I'm not particularly educated in this area of Mathematics.
Proof by contradiction consists of assuming a statement $P$ is false, and then reach a contradiction thus allowing us to conclude that $P$ must be true. Such line of reasoning seems to be using the Law of the Excluded Middle, that is, $P lor neg P$ is a tautology.
Wouldn't assuming said law lead to some problems. As an example, it has been proven that if ZFC is consistent, then both ZFC$+$CH and ZFC$+neg$CH are also consistent. Thus, by LEM, there are only two possible options:
1) CH is true, but unprovable within ZFC.
2) $neg$CH is true, but unprovable within ZFC.
Suppose for a second that the first option was correct. Since $neg$ CH is consistent with ZFC, the axiomatic system ZFC$+ neg$CH contains no contradictions. However CH being true does imply that ZFC$+ neg$CH has a contradiction. The second option being true leads to the same result.
What am I missing?
I would truly appreciate any help/thoughts.
logic soft-question
logic soft-question
edited Apr 30 at 20:55
Leo
asked Apr 30 at 20:38
LeoLeo
837517
837517
1
$begingroup$
It doesn't make sense to say CH is true in ZFC. CH is true in some models of ZFC, and it's false in other models of ZFC. (Assuming there are any models of ZFC...)
$endgroup$
– Derek Elkins
Apr 30 at 20:45
$begingroup$
what do you exactly mean by "true in ZFC"?
$endgroup$
– Zuhair
Apr 30 at 20:45
$begingroup$
You have in the title a very broad subject, and the body of your Question narrows this only in a fairly confused way. How would you define "CH is true in ZFC"? It might be more productive to begin your investigations on grounds where you have a clear understanding, rather than by combining several topics that are muddled.
$endgroup$
– hardmath
Apr 30 at 20:45
$begingroup$
I meant to say that 'CH is true along the axioms of ZFC'. I'll edit my question. I'm sorry, I'm not familiar with the correct terminology.
$endgroup$
– Leo
Apr 30 at 20:51
$begingroup$
You may be interesting in reading this question and an excellent response: math.stackexchange.com/questions/1746563/…
$endgroup$
– Vasya
Apr 30 at 20:51
|
show 4 more comments
1
$begingroup$
It doesn't make sense to say CH is true in ZFC. CH is true in some models of ZFC, and it's false in other models of ZFC. (Assuming there are any models of ZFC...)
$endgroup$
– Derek Elkins
Apr 30 at 20:45
$begingroup$
what do you exactly mean by "true in ZFC"?
$endgroup$
– Zuhair
Apr 30 at 20:45
$begingroup$
You have in the title a very broad subject, and the body of your Question narrows this only in a fairly confused way. How would you define "CH is true in ZFC"? It might be more productive to begin your investigations on grounds where you have a clear understanding, rather than by combining several topics that are muddled.
$endgroup$
– hardmath
Apr 30 at 20:45
$begingroup$
I meant to say that 'CH is true along the axioms of ZFC'. I'll edit my question. I'm sorry, I'm not familiar with the correct terminology.
$endgroup$
– Leo
Apr 30 at 20:51
$begingroup$
You may be interesting in reading this question and an excellent response: math.stackexchange.com/questions/1746563/…
$endgroup$
– Vasya
Apr 30 at 20:51
1
1
$begingroup$
It doesn't make sense to say CH is true in ZFC. CH is true in some models of ZFC, and it's false in other models of ZFC. (Assuming there are any models of ZFC...)
$endgroup$
– Derek Elkins
Apr 30 at 20:45
$begingroup$
It doesn't make sense to say CH is true in ZFC. CH is true in some models of ZFC, and it's false in other models of ZFC. (Assuming there are any models of ZFC...)
$endgroup$
– Derek Elkins
Apr 30 at 20:45
$begingroup$
what do you exactly mean by "true in ZFC"?
$endgroup$
– Zuhair
Apr 30 at 20:45
$begingroup$
what do you exactly mean by "true in ZFC"?
$endgroup$
– Zuhair
Apr 30 at 20:45
$begingroup$
You have in the title a very broad subject, and the body of your Question narrows this only in a fairly confused way. How would you define "CH is true in ZFC"? It might be more productive to begin your investigations on grounds where you have a clear understanding, rather than by combining several topics that are muddled.
$endgroup$
– hardmath
Apr 30 at 20:45
$begingroup$
You have in the title a very broad subject, and the body of your Question narrows this only in a fairly confused way. How would you define "CH is true in ZFC"? It might be more productive to begin your investigations on grounds where you have a clear understanding, rather than by combining several topics that are muddled.
$endgroup$
– hardmath
Apr 30 at 20:45
$begingroup$
I meant to say that 'CH is true along the axioms of ZFC'. I'll edit my question. I'm sorry, I'm not familiar with the correct terminology.
$endgroup$
– Leo
Apr 30 at 20:51
$begingroup$
I meant to say that 'CH is true along the axioms of ZFC'. I'll edit my question. I'm sorry, I'm not familiar with the correct terminology.
$endgroup$
– Leo
Apr 30 at 20:51
$begingroup$
You may be interesting in reading this question and an excellent response: math.stackexchange.com/questions/1746563/…
$endgroup$
– Vasya
Apr 30 at 20:51
$begingroup$
You may be interesting in reading this question and an excellent response: math.stackexchange.com/questions/1746563/…
$endgroup$
– Vasya
Apr 30 at 20:51
|
show 4 more comments
3 Answers
3
active
oldest
votes
$begingroup$
Your confusion is in conflating the truth of a set of axioms with their consistency. I'll assume ZFC is consistent throughout this explanation (that's not known, but it's assumed in the undecidability result you stated).
Let $diamond p$ denote "$p$ is consistent" and $square p$ denote "$p$ is provable" viz. modal logic (I'm tweaking its concepts slightly for the present context). Also, let $c,,z$ respectively denote the CH and ZFC. From the law of the excluded middle $clorneg c$ we deduce $zto((zland c)lor(zlandneg c))$, and the undecidability of $c$ in $z$ means that $(diamond(zland c))land(diamond(zlandneg c))$. But these results are not inconsistent. In particular, $zland c$ does not imply $square(zland c)$, and hence does not contradict $diamond(zlandneg c)$.
In particular, a general instance of the law of the excluded middle, $plorneg p$, doesn't imply $(square p)lor(square(neg p))$. Similarly, the law of non-contradiction $neg(plandneg p)$ doesn't imply $neg((diamond p)land(diamond(neg p)))$.
Just to relate all this to something you said earlier:
Proof by contradiction consists of assuming a statement $P$ is false,
and then reach a contradiction thus allowing us to conclude that $P$
must be true.
An intuitionistic logician, who rejects the law of the excluded middle, would instead say you assume some statement is true, reach a contradiction, and thus conclude the statement was false. In other words, $(qtobot)toneg q$. The case $q:=neg p$ gives $(neg ptobot)tonegneg p$, which if $p=negneg p$ simplifies to $(neg ptobot)to p$ as you intended. This simplification follows from the law of the excluded middle, but fails in intuitionistic logic. One way to understand this is that intuitionistic logic tracks provability rather than truth (this isn't necessarily how to read it, but it gives the right logical structure). In other words, $plorneg p$ fails in intuitionistic logic because $(square p)lor(square(neg p))$ fails in classical logic.
$endgroup$
$begingroup$
So is it possible for $P land neg P$ to be true in a consistent axiomatic system? I'm saying this since you claim $z land c$ doesn't contradict $z land neg c$ being consistent.
$endgroup$
– Leo
Apr 30 at 22:07
$begingroup$
Oh wait. $z land neg c$ being consistent doesn't even imply $z land neg c$, right?
$endgroup$
– Leo
Apr 30 at 22:12
$begingroup$
@Leo A consistent system is, by definition, one in which $Plandneg P$ is never true. But you work in out in your second comment; yes, a statement doesn't have to be true just because it's consistent.
$endgroup$
– J.G.
May 1 at 5:06
add a comment |
$begingroup$
Usually when it is said that a "sentence $s$ is true of some theory $T$" it is meant that $s$ is satisfied in some particular model of $T$ that is considered as a standard model of $T$, that is a model that most captures the informal concept the formal system is about.
To say that $s$ is true in $T$ does imply that $T + s$ is consistent! But it doesn't necessarily imply that $T + neg s $ is inconsistent at all. To say that $T + neg s$ is inconsistent is to say that $s$ is not satisfied in any model of $T$ and not just the standard one. The picture should be clear by now, since $s$ being satisfied in a particular model of $T$ doesn't at all imply that $neg s$ cannot be satisfied in another model of $T$.
$endgroup$
1
$begingroup$
I guess I now have to study models. I'm not even sure what they are.
$endgroup$
– Leo
Apr 30 at 21:06
2
$begingroup$
well models are the semantic counterpart of the formal system. Yes, you need to study them, otherwise it would be difficult to understand this situation.
$endgroup$
– Zuhair
Apr 30 at 21:07
add a comment |
$begingroup$
Let me make it very clear: Truth is always relative to a structure/model. Remember, every first-order sentence over a language $L$ can be true in some $L$-structure but false in another. For a trivial example "$∀x,y ( x·y = y·x )$" is true in every abelian group but false in every non-abelian group. Likewise "CH is true" is meaningless unless we specify explicitly or implicitly the structure we are talking about. Once you precisely specify this, all the problems will go away. Historically, some philosophers have made this same mistake of confusing syntax with semantics.
In fact, the independence of CH directly implies via Godel's completeness theorem that there is a model of ZFC in which CH is true, as well as a model of ZFC in which CH is false. There is simply no contradiction because those are two different models!
Some people may use "true in ZFC" to mean "provable by ZFC", but that would then mean "true in all models of ZFC", which certainly does not hold for CH. Also, if you come across statements that Godel's incompleteness theorems are about "true but unprovable sentences", note that there is still a model we are talking about, namely the standard model $mathbbN$ of PA. For more on that see this post.
$endgroup$
add a comment |
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3 Answers
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active
oldest
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3 Answers
3
active
oldest
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active
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votes
active
oldest
votes
$begingroup$
Your confusion is in conflating the truth of a set of axioms with their consistency. I'll assume ZFC is consistent throughout this explanation (that's not known, but it's assumed in the undecidability result you stated).
Let $diamond p$ denote "$p$ is consistent" and $square p$ denote "$p$ is provable" viz. modal logic (I'm tweaking its concepts slightly for the present context). Also, let $c,,z$ respectively denote the CH and ZFC. From the law of the excluded middle $clorneg c$ we deduce $zto((zland c)lor(zlandneg c))$, and the undecidability of $c$ in $z$ means that $(diamond(zland c))land(diamond(zlandneg c))$. But these results are not inconsistent. In particular, $zland c$ does not imply $square(zland c)$, and hence does not contradict $diamond(zlandneg c)$.
In particular, a general instance of the law of the excluded middle, $plorneg p$, doesn't imply $(square p)lor(square(neg p))$. Similarly, the law of non-contradiction $neg(plandneg p)$ doesn't imply $neg((diamond p)land(diamond(neg p)))$.
Just to relate all this to something you said earlier:
Proof by contradiction consists of assuming a statement $P$ is false,
and then reach a contradiction thus allowing us to conclude that $P$
must be true.
An intuitionistic logician, who rejects the law of the excluded middle, would instead say you assume some statement is true, reach a contradiction, and thus conclude the statement was false. In other words, $(qtobot)toneg q$. The case $q:=neg p$ gives $(neg ptobot)tonegneg p$, which if $p=negneg p$ simplifies to $(neg ptobot)to p$ as you intended. This simplification follows from the law of the excluded middle, but fails in intuitionistic logic. One way to understand this is that intuitionistic logic tracks provability rather than truth (this isn't necessarily how to read it, but it gives the right logical structure). In other words, $plorneg p$ fails in intuitionistic logic because $(square p)lor(square(neg p))$ fails in classical logic.
$endgroup$
$begingroup$
So is it possible for $P land neg P$ to be true in a consistent axiomatic system? I'm saying this since you claim $z land c$ doesn't contradict $z land neg c$ being consistent.
$endgroup$
– Leo
Apr 30 at 22:07
$begingroup$
Oh wait. $z land neg c$ being consistent doesn't even imply $z land neg c$, right?
$endgroup$
– Leo
Apr 30 at 22:12
$begingroup$
@Leo A consistent system is, by definition, one in which $Plandneg P$ is never true. But you work in out in your second comment; yes, a statement doesn't have to be true just because it's consistent.
$endgroup$
– J.G.
May 1 at 5:06
add a comment |
$begingroup$
Your confusion is in conflating the truth of a set of axioms with their consistency. I'll assume ZFC is consistent throughout this explanation (that's not known, but it's assumed in the undecidability result you stated).
Let $diamond p$ denote "$p$ is consistent" and $square p$ denote "$p$ is provable" viz. modal logic (I'm tweaking its concepts slightly for the present context). Also, let $c,,z$ respectively denote the CH and ZFC. From the law of the excluded middle $clorneg c$ we deduce $zto((zland c)lor(zlandneg c))$, and the undecidability of $c$ in $z$ means that $(diamond(zland c))land(diamond(zlandneg c))$. But these results are not inconsistent. In particular, $zland c$ does not imply $square(zland c)$, and hence does not contradict $diamond(zlandneg c)$.
In particular, a general instance of the law of the excluded middle, $plorneg p$, doesn't imply $(square p)lor(square(neg p))$. Similarly, the law of non-contradiction $neg(plandneg p)$ doesn't imply $neg((diamond p)land(diamond(neg p)))$.
Just to relate all this to something you said earlier:
Proof by contradiction consists of assuming a statement $P$ is false,
and then reach a contradiction thus allowing us to conclude that $P$
must be true.
An intuitionistic logician, who rejects the law of the excluded middle, would instead say you assume some statement is true, reach a contradiction, and thus conclude the statement was false. In other words, $(qtobot)toneg q$. The case $q:=neg p$ gives $(neg ptobot)tonegneg p$, which if $p=negneg p$ simplifies to $(neg ptobot)to p$ as you intended. This simplification follows from the law of the excluded middle, but fails in intuitionistic logic. One way to understand this is that intuitionistic logic tracks provability rather than truth (this isn't necessarily how to read it, but it gives the right logical structure). In other words, $plorneg p$ fails in intuitionistic logic because $(square p)lor(square(neg p))$ fails in classical logic.
$endgroup$
$begingroup$
So is it possible for $P land neg P$ to be true in a consistent axiomatic system? I'm saying this since you claim $z land c$ doesn't contradict $z land neg c$ being consistent.
$endgroup$
– Leo
Apr 30 at 22:07
$begingroup$
Oh wait. $z land neg c$ being consistent doesn't even imply $z land neg c$, right?
$endgroup$
– Leo
Apr 30 at 22:12
$begingroup$
@Leo A consistent system is, by definition, one in which $Plandneg P$ is never true. But you work in out in your second comment; yes, a statement doesn't have to be true just because it's consistent.
$endgroup$
– J.G.
May 1 at 5:06
add a comment |
$begingroup$
Your confusion is in conflating the truth of a set of axioms with their consistency. I'll assume ZFC is consistent throughout this explanation (that's not known, but it's assumed in the undecidability result you stated).
Let $diamond p$ denote "$p$ is consistent" and $square p$ denote "$p$ is provable" viz. modal logic (I'm tweaking its concepts slightly for the present context). Also, let $c,,z$ respectively denote the CH and ZFC. From the law of the excluded middle $clorneg c$ we deduce $zto((zland c)lor(zlandneg c))$, and the undecidability of $c$ in $z$ means that $(diamond(zland c))land(diamond(zlandneg c))$. But these results are not inconsistent. In particular, $zland c$ does not imply $square(zland c)$, and hence does not contradict $diamond(zlandneg c)$.
In particular, a general instance of the law of the excluded middle, $plorneg p$, doesn't imply $(square p)lor(square(neg p))$. Similarly, the law of non-contradiction $neg(plandneg p)$ doesn't imply $neg((diamond p)land(diamond(neg p)))$.
Just to relate all this to something you said earlier:
Proof by contradiction consists of assuming a statement $P$ is false,
and then reach a contradiction thus allowing us to conclude that $P$
must be true.
An intuitionistic logician, who rejects the law of the excluded middle, would instead say you assume some statement is true, reach a contradiction, and thus conclude the statement was false. In other words, $(qtobot)toneg q$. The case $q:=neg p$ gives $(neg ptobot)tonegneg p$, which if $p=negneg p$ simplifies to $(neg ptobot)to p$ as you intended. This simplification follows from the law of the excluded middle, but fails in intuitionistic logic. One way to understand this is that intuitionistic logic tracks provability rather than truth (this isn't necessarily how to read it, but it gives the right logical structure). In other words, $plorneg p$ fails in intuitionistic logic because $(square p)lor(square(neg p))$ fails in classical logic.
$endgroup$
Your confusion is in conflating the truth of a set of axioms with their consistency. I'll assume ZFC is consistent throughout this explanation (that's not known, but it's assumed in the undecidability result you stated).
Let $diamond p$ denote "$p$ is consistent" and $square p$ denote "$p$ is provable" viz. modal logic (I'm tweaking its concepts slightly for the present context). Also, let $c,,z$ respectively denote the CH and ZFC. From the law of the excluded middle $clorneg c$ we deduce $zto((zland c)lor(zlandneg c))$, and the undecidability of $c$ in $z$ means that $(diamond(zland c))land(diamond(zlandneg c))$. But these results are not inconsistent. In particular, $zland c$ does not imply $square(zland c)$, and hence does not contradict $diamond(zlandneg c)$.
In particular, a general instance of the law of the excluded middle, $plorneg p$, doesn't imply $(square p)lor(square(neg p))$. Similarly, the law of non-contradiction $neg(plandneg p)$ doesn't imply $neg((diamond p)land(diamond(neg p)))$.
Just to relate all this to something you said earlier:
Proof by contradiction consists of assuming a statement $P$ is false,
and then reach a contradiction thus allowing us to conclude that $P$
must be true.
An intuitionistic logician, who rejects the law of the excluded middle, would instead say you assume some statement is true, reach a contradiction, and thus conclude the statement was false. In other words, $(qtobot)toneg q$. The case $q:=neg p$ gives $(neg ptobot)tonegneg p$, which if $p=negneg p$ simplifies to $(neg ptobot)to p$ as you intended. This simplification follows from the law of the excluded middle, but fails in intuitionistic logic. One way to understand this is that intuitionistic logic tracks provability rather than truth (this isn't necessarily how to read it, but it gives the right logical structure). In other words, $plorneg p$ fails in intuitionistic logic because $(square p)lor(square(neg p))$ fails in classical logic.
answered Apr 30 at 21:08
J.G.J.G.
35.6k23554
35.6k23554
$begingroup$
So is it possible for $P land neg P$ to be true in a consistent axiomatic system? I'm saying this since you claim $z land c$ doesn't contradict $z land neg c$ being consistent.
$endgroup$
– Leo
Apr 30 at 22:07
$begingroup$
Oh wait. $z land neg c$ being consistent doesn't even imply $z land neg c$, right?
$endgroup$
– Leo
Apr 30 at 22:12
$begingroup$
@Leo A consistent system is, by definition, one in which $Plandneg P$ is never true. But you work in out in your second comment; yes, a statement doesn't have to be true just because it's consistent.
$endgroup$
– J.G.
May 1 at 5:06
add a comment |
$begingroup$
So is it possible for $P land neg P$ to be true in a consistent axiomatic system? I'm saying this since you claim $z land c$ doesn't contradict $z land neg c$ being consistent.
$endgroup$
– Leo
Apr 30 at 22:07
$begingroup$
Oh wait. $z land neg c$ being consistent doesn't even imply $z land neg c$, right?
$endgroup$
– Leo
Apr 30 at 22:12
$begingroup$
@Leo A consistent system is, by definition, one in which $Plandneg P$ is never true. But you work in out in your second comment; yes, a statement doesn't have to be true just because it's consistent.
$endgroup$
– J.G.
May 1 at 5:06
$begingroup$
So is it possible for $P land neg P$ to be true in a consistent axiomatic system? I'm saying this since you claim $z land c$ doesn't contradict $z land neg c$ being consistent.
$endgroup$
– Leo
Apr 30 at 22:07
$begingroup$
So is it possible for $P land neg P$ to be true in a consistent axiomatic system? I'm saying this since you claim $z land c$ doesn't contradict $z land neg c$ being consistent.
$endgroup$
– Leo
Apr 30 at 22:07
$begingroup$
Oh wait. $z land neg c$ being consistent doesn't even imply $z land neg c$, right?
$endgroup$
– Leo
Apr 30 at 22:12
$begingroup$
Oh wait. $z land neg c$ being consistent doesn't even imply $z land neg c$, right?
$endgroup$
– Leo
Apr 30 at 22:12
$begingroup$
@Leo A consistent system is, by definition, one in which $Plandneg P$ is never true. But you work in out in your second comment; yes, a statement doesn't have to be true just because it's consistent.
$endgroup$
– J.G.
May 1 at 5:06
$begingroup$
@Leo A consistent system is, by definition, one in which $Plandneg P$ is never true. But you work in out in your second comment; yes, a statement doesn't have to be true just because it's consistent.
$endgroup$
– J.G.
May 1 at 5:06
add a comment |
$begingroup$
Usually when it is said that a "sentence $s$ is true of some theory $T$" it is meant that $s$ is satisfied in some particular model of $T$ that is considered as a standard model of $T$, that is a model that most captures the informal concept the formal system is about.
To say that $s$ is true in $T$ does imply that $T + s$ is consistent! But it doesn't necessarily imply that $T + neg s $ is inconsistent at all. To say that $T + neg s$ is inconsistent is to say that $s$ is not satisfied in any model of $T$ and not just the standard one. The picture should be clear by now, since $s$ being satisfied in a particular model of $T$ doesn't at all imply that $neg s$ cannot be satisfied in another model of $T$.
$endgroup$
1
$begingroup$
I guess I now have to study models. I'm not even sure what they are.
$endgroup$
– Leo
Apr 30 at 21:06
2
$begingroup$
well models are the semantic counterpart of the formal system. Yes, you need to study them, otherwise it would be difficult to understand this situation.
$endgroup$
– Zuhair
Apr 30 at 21:07
add a comment |
$begingroup$
Usually when it is said that a "sentence $s$ is true of some theory $T$" it is meant that $s$ is satisfied in some particular model of $T$ that is considered as a standard model of $T$, that is a model that most captures the informal concept the formal system is about.
To say that $s$ is true in $T$ does imply that $T + s$ is consistent! But it doesn't necessarily imply that $T + neg s $ is inconsistent at all. To say that $T + neg s$ is inconsistent is to say that $s$ is not satisfied in any model of $T$ and not just the standard one. The picture should be clear by now, since $s$ being satisfied in a particular model of $T$ doesn't at all imply that $neg s$ cannot be satisfied in another model of $T$.
$endgroup$
1
$begingroup$
I guess I now have to study models. I'm not even sure what they are.
$endgroup$
– Leo
Apr 30 at 21:06
2
$begingroup$
well models are the semantic counterpart of the formal system. Yes, you need to study them, otherwise it would be difficult to understand this situation.
$endgroup$
– Zuhair
Apr 30 at 21:07
add a comment |
$begingroup$
Usually when it is said that a "sentence $s$ is true of some theory $T$" it is meant that $s$ is satisfied in some particular model of $T$ that is considered as a standard model of $T$, that is a model that most captures the informal concept the formal system is about.
To say that $s$ is true in $T$ does imply that $T + s$ is consistent! But it doesn't necessarily imply that $T + neg s $ is inconsistent at all. To say that $T + neg s$ is inconsistent is to say that $s$ is not satisfied in any model of $T$ and not just the standard one. The picture should be clear by now, since $s$ being satisfied in a particular model of $T$ doesn't at all imply that $neg s$ cannot be satisfied in another model of $T$.
$endgroup$
Usually when it is said that a "sentence $s$ is true of some theory $T$" it is meant that $s$ is satisfied in some particular model of $T$ that is considered as a standard model of $T$, that is a model that most captures the informal concept the formal system is about.
To say that $s$ is true in $T$ does imply that $T + s$ is consistent! But it doesn't necessarily imply that $T + neg s $ is inconsistent at all. To say that $T + neg s$ is inconsistent is to say that $s$ is not satisfied in any model of $T$ and not just the standard one. The picture should be clear by now, since $s$ being satisfied in a particular model of $T$ doesn't at all imply that $neg s$ cannot be satisfied in another model of $T$.
answered Apr 30 at 21:02
ZuhairZuhair
435212
435212
1
$begingroup$
I guess I now have to study models. I'm not even sure what they are.
$endgroup$
– Leo
Apr 30 at 21:06
2
$begingroup$
well models are the semantic counterpart of the formal system. Yes, you need to study them, otherwise it would be difficult to understand this situation.
$endgroup$
– Zuhair
Apr 30 at 21:07
add a comment |
1
$begingroup$
I guess I now have to study models. I'm not even sure what they are.
$endgroup$
– Leo
Apr 30 at 21:06
2
$begingroup$
well models are the semantic counterpart of the formal system. Yes, you need to study them, otherwise it would be difficult to understand this situation.
$endgroup$
– Zuhair
Apr 30 at 21:07
1
1
$begingroup$
I guess I now have to study models. I'm not even sure what they are.
$endgroup$
– Leo
Apr 30 at 21:06
$begingroup$
I guess I now have to study models. I'm not even sure what they are.
$endgroup$
– Leo
Apr 30 at 21:06
2
2
$begingroup$
well models are the semantic counterpart of the formal system. Yes, you need to study them, otherwise it would be difficult to understand this situation.
$endgroup$
– Zuhair
Apr 30 at 21:07
$begingroup$
well models are the semantic counterpart of the formal system. Yes, you need to study them, otherwise it would be difficult to understand this situation.
$endgroup$
– Zuhair
Apr 30 at 21:07
add a comment |
$begingroup$
Let me make it very clear: Truth is always relative to a structure/model. Remember, every first-order sentence over a language $L$ can be true in some $L$-structure but false in another. For a trivial example "$∀x,y ( x·y = y·x )$" is true in every abelian group but false in every non-abelian group. Likewise "CH is true" is meaningless unless we specify explicitly or implicitly the structure we are talking about. Once you precisely specify this, all the problems will go away. Historically, some philosophers have made this same mistake of confusing syntax with semantics.
In fact, the independence of CH directly implies via Godel's completeness theorem that there is a model of ZFC in which CH is true, as well as a model of ZFC in which CH is false. There is simply no contradiction because those are two different models!
Some people may use "true in ZFC" to mean "provable by ZFC", but that would then mean "true in all models of ZFC", which certainly does not hold for CH. Also, if you come across statements that Godel's incompleteness theorems are about "true but unprovable sentences", note that there is still a model we are talking about, namely the standard model $mathbbN$ of PA. For more on that see this post.
$endgroup$
add a comment |
$begingroup$
Let me make it very clear: Truth is always relative to a structure/model. Remember, every first-order sentence over a language $L$ can be true in some $L$-structure but false in another. For a trivial example "$∀x,y ( x·y = y·x )$" is true in every abelian group but false in every non-abelian group. Likewise "CH is true" is meaningless unless we specify explicitly or implicitly the structure we are talking about. Once you precisely specify this, all the problems will go away. Historically, some philosophers have made this same mistake of confusing syntax with semantics.
In fact, the independence of CH directly implies via Godel's completeness theorem that there is a model of ZFC in which CH is true, as well as a model of ZFC in which CH is false. There is simply no contradiction because those are two different models!
Some people may use "true in ZFC" to mean "provable by ZFC", but that would then mean "true in all models of ZFC", which certainly does not hold for CH. Also, if you come across statements that Godel's incompleteness theorems are about "true but unprovable sentences", note that there is still a model we are talking about, namely the standard model $mathbbN$ of PA. For more on that see this post.
$endgroup$
add a comment |
$begingroup$
Let me make it very clear: Truth is always relative to a structure/model. Remember, every first-order sentence over a language $L$ can be true in some $L$-structure but false in another. For a trivial example "$∀x,y ( x·y = y·x )$" is true in every abelian group but false in every non-abelian group. Likewise "CH is true" is meaningless unless we specify explicitly or implicitly the structure we are talking about. Once you precisely specify this, all the problems will go away. Historically, some philosophers have made this same mistake of confusing syntax with semantics.
In fact, the independence of CH directly implies via Godel's completeness theorem that there is a model of ZFC in which CH is true, as well as a model of ZFC in which CH is false. There is simply no contradiction because those are two different models!
Some people may use "true in ZFC" to mean "provable by ZFC", but that would then mean "true in all models of ZFC", which certainly does not hold for CH. Also, if you come across statements that Godel's incompleteness theorems are about "true but unprovable sentences", note that there is still a model we are talking about, namely the standard model $mathbbN$ of PA. For more on that see this post.
$endgroup$
Let me make it very clear: Truth is always relative to a structure/model. Remember, every first-order sentence over a language $L$ can be true in some $L$-structure but false in another. For a trivial example "$∀x,y ( x·y = y·x )$" is true in every abelian group but false in every non-abelian group. Likewise "CH is true" is meaningless unless we specify explicitly or implicitly the structure we are talking about. Once you precisely specify this, all the problems will go away. Historically, some philosophers have made this same mistake of confusing syntax with semantics.
In fact, the independence of CH directly implies via Godel's completeness theorem that there is a model of ZFC in which CH is true, as well as a model of ZFC in which CH is false. There is simply no contradiction because those are two different models!
Some people may use "true in ZFC" to mean "provable by ZFC", but that would then mean "true in all models of ZFC", which certainly does not hold for CH. Also, if you come across statements that Godel's incompleteness theorems are about "true but unprovable sentences", note that there is still a model we are talking about, namely the standard model $mathbbN$ of PA. For more on that see this post.
answered May 1 at 2:18
user21820user21820
40.5k544163
40.5k544163
add a comment |
add a comment |
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$begingroup$
It doesn't make sense to say CH is true in ZFC. CH is true in some models of ZFC, and it's false in other models of ZFC. (Assuming there are any models of ZFC...)
$endgroup$
– Derek Elkins
Apr 30 at 20:45
$begingroup$
what do you exactly mean by "true in ZFC"?
$endgroup$
– Zuhair
Apr 30 at 20:45
$begingroup$
You have in the title a very broad subject, and the body of your Question narrows this only in a fairly confused way. How would you define "CH is true in ZFC"? It might be more productive to begin your investigations on grounds where you have a clear understanding, rather than by combining several topics that are muddled.
$endgroup$
– hardmath
Apr 30 at 20:45
$begingroup$
I meant to say that 'CH is true along the axioms of ZFC'. I'll edit my question. I'm sorry, I'm not familiar with the correct terminology.
$endgroup$
– Leo
Apr 30 at 20:51
$begingroup$
You may be interesting in reading this question and an excellent response: math.stackexchange.com/questions/1746563/…
$endgroup$
– Vasya
Apr 30 at 20:51