How can ZF prove relative consistency for itself?












2














This is related to my first question. In order to get what I don't get, I ll go with something much more specific here.



It is well known that $ZF vdash Con(ZF) longrightarrow Con(ZF + AF)$. The way I know to prove it uses :



Lemma 1 : If $T$ and $S$ are two theories in set theory language, $M$ is a class (i.e. a formula $phi(x)$ tell if $x$ is in $M$ or not) and $T vdash mbox{"}M$ is a model for $S$", then $Con(T) vdash Con(S)$



So we have to build a class, say $V=cup_{alpha in ORD} V_{alpha}$ and then $ZF$ is suppose to demonstrate, for instance, that $V$ satisfies comprehension axiom scheme.



We can use :



Lemma 2 : $(forall x in M$, $P(x) subset M) longrightarrow$ comprehension scheme is true in $M$



With Lemma 2, the proof is straightforward.
My problem is that I do not know why lemma 2 would be a ZF theorem. I don't even know how it could be written in ZF language, since there is an infinite number of axiom to write, and since $V$ is a class, there is no formula $theta(ulcorner f urcorner)$ saying $V vDash f$



Of course, for all formulas $F$ of the comprehension scheme, we can write



$ZF vdash (forall x in M$, $P(x) subset M) longrightarrow F^M$



but if we do that for all formulas, the proof will be infinite...






logic set-theory







share|cite|improve this question
























  • Be careful with the fact that this approach to proving relative consistency takes place in the metatheory, not in the object language: math.stackexchange.com/a/1892103/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34










  • See also: math.stackexchange.com/a/1368646/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34
















2














This is related to my first question. In order to get what I don't get, I ll go with something much more specific here.



It is well known that $ZF vdash Con(ZF) longrightarrow Con(ZF + AF)$. The way I know to prove it uses :



Lemma 1 : If $T$ and $S$ are two theories in set theory language, $M$ is a class (i.e. a formula $phi(x)$ tell if $x$ is in $M$ or not) and $T vdash mbox{"}M$ is a model for $S$", then $Con(T) vdash Con(S)$



So we have to build a class, say $V=cup_{alpha in ORD} V_{alpha}$ and then $ZF$ is suppose to demonstrate, for instance, that $V$ satisfies comprehension axiom scheme.



We can use :



Lemma 2 : $(forall x in M$, $P(x) subset M) longrightarrow$ comprehension scheme is true in $M$



With Lemma 2, the proof is straightforward.
My problem is that I do not know why lemma 2 would be a ZF theorem. I don't even know how it could be written in ZF language, since there is an infinite number of axiom to write, and since $V$ is a class, there is no formula $theta(ulcorner f urcorner)$ saying $V vDash f$



Of course, for all formulas $F$ of the comprehension scheme, we can write



$ZF vdash (forall x in M$, $P(x) subset M) longrightarrow F^M$



but if we do that for all formulas, the proof will be infinite...






logic set-theory







share|cite|improve this question
























  • Be careful with the fact that this approach to proving relative consistency takes place in the metatheory, not in the object language: math.stackexchange.com/a/1892103/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34










  • See also: math.stackexchange.com/a/1368646/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34














2












2








2







This is related to my first question. In order to get what I don't get, I ll go with something much more specific here.



It is well known that $ZF vdash Con(ZF) longrightarrow Con(ZF + AF)$. The way I know to prove it uses :



Lemma 1 : If $T$ and $S$ are two theories in set theory language, $M$ is a class (i.e. a formula $phi(x)$ tell if $x$ is in $M$ or not) and $T vdash mbox{"}M$ is a model for $S$", then $Con(T) vdash Con(S)$



So we have to build a class, say $V=cup_{alpha in ORD} V_{alpha}$ and then $ZF$ is suppose to demonstrate, for instance, that $V$ satisfies comprehension axiom scheme.



We can use :



Lemma 2 : $(forall x in M$, $P(x) subset M) longrightarrow$ comprehension scheme is true in $M$



With Lemma 2, the proof is straightforward.
My problem is that I do not know why lemma 2 would be a ZF theorem. I don't even know how it could be written in ZF language, since there is an infinite number of axiom to write, and since $V$ is a class, there is no formula $theta(ulcorner f urcorner)$ saying $V vDash f$



Of course, for all formulas $F$ of the comprehension scheme, we can write



$ZF vdash (forall x in M$, $P(x) subset M) longrightarrow F^M$



but if we do that for all formulas, the proof will be infinite...






logic set-theory







share|cite|improve this question















This is related to my first question. In order to get what I don't get, I ll go with something much more specific here.



It is well known that $ZF vdash Con(ZF) longrightarrow Con(ZF + AF)$. The way I know to prove it uses :



Lemma 1 : If $T$ and $S$ are two theories in set theory language, $M$ is a class (i.e. a formula $phi(x)$ tell if $x$ is in $M$ or not) and $T vdash mbox{"}M$ is a model for $S$", then $Con(T) vdash Con(S)$



So we have to build a class, say $V=cup_{alpha in ORD} V_{alpha}$ and then $ZF$ is suppose to demonstrate, for instance, that $V$ satisfies comprehension axiom scheme.



We can use :



Lemma 2 : $(forall x in M$, $P(x) subset M) longrightarrow$ comprehension scheme is true in $M$



With Lemma 2, the proof is straightforward.
My problem is that I do not know why lemma 2 would be a ZF theorem. I don't even know how it could be written in ZF language, since there is an infinite number of axiom to write, and since $V$ is a class, there is no formula $theta(ulcorner f urcorner)$ saying $V vDash f$



Of course, for all formulas $F$ of the comprehension scheme, we can write



$ZF vdash (forall x in M$, $P(x) subset M) longrightarrow F^M$



but if we do that for all formulas, the proof will be infinite...






logic set-theory




logic set-theory






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 26 at 4:32









Hans Hüttel

3,1872921




3,1872921










asked Mar 22 '11 at 16:26









Archimondain

63639




63639












  • Be careful with the fact that this approach to proving relative consistency takes place in the metatheory, not in the object language: math.stackexchange.com/a/1892103/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34










  • See also: math.stackexchange.com/a/1368646/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34


















  • Be careful with the fact that this approach to proving relative consistency takes place in the metatheory, not in the object language: math.stackexchange.com/a/1892103/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34










  • See also: math.stackexchange.com/a/1368646/75867
    – Ioannis Filippidis
    Dec 10 '16 at 9:34
















Be careful with the fact that this approach to proving relative consistency takes place in the metatheory, not in the object language: math.stackexchange.com/a/1892103/75867
– Ioannis Filippidis
Dec 10 '16 at 9:34




Be careful with the fact that this approach to proving relative consistency takes place in the metatheory, not in the object language: math.stackexchange.com/a/1892103/75867
– Ioannis Filippidis
Dec 10 '16 at 9:34












See also: math.stackexchange.com/a/1368646/75867
– Ioannis Filippidis
Dec 10 '16 at 9:34




See also: math.stackexchange.com/a/1368646/75867
– Ioannis Filippidis
Dec 10 '16 at 9:34










1 Answer
1






active

oldest

votes


















2














(In what follows, I am assuming that ZF consists of Extensionality, Pairing, Union, Comprehension, Replacement, Infinity and Power Set, and that AF is the Axiom of Foundation.)



I think that the somewhat formal way of looking at all of this is to basically work by contrapositive. Working in ZF, assume $neg mathrm{Con} ( text{ZF+AF} )$. Then there is a proof of some obvious contradiction, say $( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$, in ZF+AF. Furthermore, ZF is able to code proofs/formulas as sets/natural numbers in a standard manner. This proof can only consist of finitely many axioms from ZF+AF, say $phi_1 , ldots , phi_n$. Also in ZF we can construct the class of well-founded sets, denoted (as in Kunen) by $mathbf{WF}$, and in ZF we can prove that $mathbf{WF} models phi_1 wedge cdots wedge phi_n$, and therefore it must be that ''$mathbf{WF} models ( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$'' is a theorem of $text{ZF} + neg mathrm{Con} ( text{ZF+ AF} )$. Formally, using relativisation, we have that $$text{ZF} + neg mathrm{Con} ( text{ZF+AF} ) vdash ( forall x in mathbf{WF} ) ( x = x ) wedge neg ( forall x in mathbf{WF} ) ( x = x ),$$ and thus $text{ZF} vdash neg mathrm{Con} ( text{ZF+AF} ) rightarrow neg mathrm{Con} ( text{ZF} )$.






share|cite|improve this answer





















  • Awesome ! This answers everything I was wondering. thx a lot. So despite the fact that this is wrong to say $ZF vdash mbox{"}WF vDash ZFmbox{"}$, this is true to say $ZF vdash Con(ZF) rightarrow Con(ZF+AF)$
    – Archimondain
    Mar 22 '11 at 17:54












  • Actually, $neg Con(ZF + AF)$ says there is a proof of some obvios contradiction, but why would this proof be standard ? If it is not, how to proove $neg Con(ZF)$ ?
    – Archimondain
    Mar 23 '11 at 9:47











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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









2














(In what follows, I am assuming that ZF consists of Extensionality, Pairing, Union, Comprehension, Replacement, Infinity and Power Set, and that AF is the Axiom of Foundation.)



I think that the somewhat formal way of looking at all of this is to basically work by contrapositive. Working in ZF, assume $neg mathrm{Con} ( text{ZF+AF} )$. Then there is a proof of some obvious contradiction, say $( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$, in ZF+AF. Furthermore, ZF is able to code proofs/formulas as sets/natural numbers in a standard manner. This proof can only consist of finitely many axioms from ZF+AF, say $phi_1 , ldots , phi_n$. Also in ZF we can construct the class of well-founded sets, denoted (as in Kunen) by $mathbf{WF}$, and in ZF we can prove that $mathbf{WF} models phi_1 wedge cdots wedge phi_n$, and therefore it must be that ''$mathbf{WF} models ( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$'' is a theorem of $text{ZF} + neg mathrm{Con} ( text{ZF+ AF} )$. Formally, using relativisation, we have that $$text{ZF} + neg mathrm{Con} ( text{ZF+AF} ) vdash ( forall x in mathbf{WF} ) ( x = x ) wedge neg ( forall x in mathbf{WF} ) ( x = x ),$$ and thus $text{ZF} vdash neg mathrm{Con} ( text{ZF+AF} ) rightarrow neg mathrm{Con} ( text{ZF} )$.






share|cite|improve this answer





















  • Awesome ! This answers everything I was wondering. thx a lot. So despite the fact that this is wrong to say $ZF vdash mbox{"}WF vDash ZFmbox{"}$, this is true to say $ZF vdash Con(ZF) rightarrow Con(ZF+AF)$
    – Archimondain
    Mar 22 '11 at 17:54












  • Actually, $neg Con(ZF + AF)$ says there is a proof of some obvios contradiction, but why would this proof be standard ? If it is not, how to proove $neg Con(ZF)$ ?
    – Archimondain
    Mar 23 '11 at 9:47
















2














(In what follows, I am assuming that ZF consists of Extensionality, Pairing, Union, Comprehension, Replacement, Infinity and Power Set, and that AF is the Axiom of Foundation.)



I think that the somewhat formal way of looking at all of this is to basically work by contrapositive. Working in ZF, assume $neg mathrm{Con} ( text{ZF+AF} )$. Then there is a proof of some obvious contradiction, say $( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$, in ZF+AF. Furthermore, ZF is able to code proofs/formulas as sets/natural numbers in a standard manner. This proof can only consist of finitely many axioms from ZF+AF, say $phi_1 , ldots , phi_n$. Also in ZF we can construct the class of well-founded sets, denoted (as in Kunen) by $mathbf{WF}$, and in ZF we can prove that $mathbf{WF} models phi_1 wedge cdots wedge phi_n$, and therefore it must be that ''$mathbf{WF} models ( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$'' is a theorem of $text{ZF} + neg mathrm{Con} ( text{ZF+ AF} )$. Formally, using relativisation, we have that $$text{ZF} + neg mathrm{Con} ( text{ZF+AF} ) vdash ( forall x in mathbf{WF} ) ( x = x ) wedge neg ( forall x in mathbf{WF} ) ( x = x ),$$ and thus $text{ZF} vdash neg mathrm{Con} ( text{ZF+AF} ) rightarrow neg mathrm{Con} ( text{ZF} )$.






share|cite|improve this answer





















  • Awesome ! This answers everything I was wondering. thx a lot. So despite the fact that this is wrong to say $ZF vdash mbox{"}WF vDash ZFmbox{"}$, this is true to say $ZF vdash Con(ZF) rightarrow Con(ZF+AF)$
    – Archimondain
    Mar 22 '11 at 17:54












  • Actually, $neg Con(ZF + AF)$ says there is a proof of some obvios contradiction, but why would this proof be standard ? If it is not, how to proove $neg Con(ZF)$ ?
    – Archimondain
    Mar 23 '11 at 9:47














2












2








2






(In what follows, I am assuming that ZF consists of Extensionality, Pairing, Union, Comprehension, Replacement, Infinity and Power Set, and that AF is the Axiom of Foundation.)



I think that the somewhat formal way of looking at all of this is to basically work by contrapositive. Working in ZF, assume $neg mathrm{Con} ( text{ZF+AF} )$. Then there is a proof of some obvious contradiction, say $( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$, in ZF+AF. Furthermore, ZF is able to code proofs/formulas as sets/natural numbers in a standard manner. This proof can only consist of finitely many axioms from ZF+AF, say $phi_1 , ldots , phi_n$. Also in ZF we can construct the class of well-founded sets, denoted (as in Kunen) by $mathbf{WF}$, and in ZF we can prove that $mathbf{WF} models phi_1 wedge cdots wedge phi_n$, and therefore it must be that ''$mathbf{WF} models ( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$'' is a theorem of $text{ZF} + neg mathrm{Con} ( text{ZF+ AF} )$. Formally, using relativisation, we have that $$text{ZF} + neg mathrm{Con} ( text{ZF+AF} ) vdash ( forall x in mathbf{WF} ) ( x = x ) wedge neg ( forall x in mathbf{WF} ) ( x = x ),$$ and thus $text{ZF} vdash neg mathrm{Con} ( text{ZF+AF} ) rightarrow neg mathrm{Con} ( text{ZF} )$.






share|cite|improve this answer












(In what follows, I am assuming that ZF consists of Extensionality, Pairing, Union, Comprehension, Replacement, Infinity and Power Set, and that AF is the Axiom of Foundation.)



I think that the somewhat formal way of looking at all of this is to basically work by contrapositive. Working in ZF, assume $neg mathrm{Con} ( text{ZF+AF} )$. Then there is a proof of some obvious contradiction, say $( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$, in ZF+AF. Furthermore, ZF is able to code proofs/formulas as sets/natural numbers in a standard manner. This proof can only consist of finitely many axioms from ZF+AF, say $phi_1 , ldots , phi_n$. Also in ZF we can construct the class of well-founded sets, denoted (as in Kunen) by $mathbf{WF}$, and in ZF we can prove that $mathbf{WF} models phi_1 wedge cdots wedge phi_n$, and therefore it must be that ''$mathbf{WF} models ( forall x ) ( x = x ) wedge neg ( forall x ) ( x = x )$'' is a theorem of $text{ZF} + neg mathrm{Con} ( text{ZF+ AF} )$. Formally, using relativisation, we have that $$text{ZF} + neg mathrm{Con} ( text{ZF+AF} ) vdash ( forall x in mathbf{WF} ) ( x = x ) wedge neg ( forall x in mathbf{WF} ) ( x = x ),$$ and thus $text{ZF} vdash neg mathrm{Con} ( text{ZF+AF} ) rightarrow neg mathrm{Con} ( text{ZF} )$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Mar 22 '11 at 17:30









user642796

44.5k559116




44.5k559116












  • Awesome ! This answers everything I was wondering. thx a lot. So despite the fact that this is wrong to say $ZF vdash mbox{"}WF vDash ZFmbox{"}$, this is true to say $ZF vdash Con(ZF) rightarrow Con(ZF+AF)$
    – Archimondain
    Mar 22 '11 at 17:54












  • Actually, $neg Con(ZF + AF)$ says there is a proof of some obvios contradiction, but why would this proof be standard ? If it is not, how to proove $neg Con(ZF)$ ?
    – Archimondain
    Mar 23 '11 at 9:47


















  • Awesome ! This answers everything I was wondering. thx a lot. So despite the fact that this is wrong to say $ZF vdash mbox{"}WF vDash ZFmbox{"}$, this is true to say $ZF vdash Con(ZF) rightarrow Con(ZF+AF)$
    – Archimondain
    Mar 22 '11 at 17:54












  • Actually, $neg Con(ZF + AF)$ says there is a proof of some obvios contradiction, but why would this proof be standard ? If it is not, how to proove $neg Con(ZF)$ ?
    – Archimondain
    Mar 23 '11 at 9:47
















Awesome ! This answers everything I was wondering. thx a lot. So despite the fact that this is wrong to say $ZF vdash mbox{"}WF vDash ZFmbox{"}$, this is true to say $ZF vdash Con(ZF) rightarrow Con(ZF+AF)$
– Archimondain
Mar 22 '11 at 17:54






Awesome ! This answers everything I was wondering. thx a lot. So despite the fact that this is wrong to say $ZF vdash mbox{"}WF vDash ZFmbox{"}$, this is true to say $ZF vdash Con(ZF) rightarrow Con(ZF+AF)$
– Archimondain
Mar 22 '11 at 17:54














Actually, $neg Con(ZF + AF)$ says there is a proof of some obvios contradiction, but why would this proof be standard ? If it is not, how to proove $neg Con(ZF)$ ?
– Archimondain
Mar 23 '11 at 9:47




Actually, $neg Con(ZF + AF)$ says there is a proof of some obvios contradiction, but why would this proof be standard ? If it is not, how to proove $neg Con(ZF)$ ?
– Archimondain
Mar 23 '11 at 9:47


















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