Mixing
A (quantifier-free) is true in standard model of PA ==> PA |- A ??
$begingroup$
Is the following statement correct ?
A is a formula in PA without a quantifier and
A is true for the standard model of arithmetic, i.e. the model
|N = (N,+,×,0,1,<)
This means: |N |= A
==>
A is proofable in PA. This means:
PA |- A
Example:
|N |= (x+y)^2 =x^2 + 2xy + y^2
===>
PA |- (x+y)^2 =x^2 + 2xy + y^2
If this statement is wrong, please give me a counterexample
cu
cx
peano-axioms
asked Jan 29 at 15:27
user508589user508589
12
$endgroup$
add a comment |
$begingroup$
Is the following statement correct ?
A is a formula in PA without a quantifier and
A is true for the standard model of arithmetic, i.e. the model
|N = (N,+,×,0,1,<)
This means: |N |= A
==>
A is proofable in PA. This means:
PA |- A
Example:
|N |= (x+y)^2 =x^2 + 2xy + y^2
===>
PA |- (x+y)^2 =x^2 + 2xy + y^2
If this statement is wrong, please give me a counterexample
cu
cx
peano-axioms
asked Jan 29 at 15:27
user508589user508589
12
$endgroup$
add a comment |
$begingroup$
Is the following statement correct ?
A is a formula in PA without a quantifier and
A is true for the standard model of arithmetic, i.e. the model
|N = (N,+,×,0,1,<)
This means: |N |= A
==>
A is proofable in PA. This means:
PA |- A
Example:
|N |= (x+y)^2 =x^2 + 2xy + y^2
===>
PA |- (x+y)^2 =x^2 + 2xy + y^2
If this statement is wrong, please give me a counterexample
cu
cx
peano-axioms
asked Jan 29 at 15:27
user508589user508589
12
$endgroup$
Is the following statement correct ?
A is a formula in PA without a quantifier and
A is true for the standard model of arithmetic, i.e. the model
|N = (N,+,×,0,1,<)
This means: |N |= A
==>
A is proofable in PA. This means:
PA |- A
Example:
|N |= (x+y)^2 =x^2 + 2xy + y^2
===>
PA |- (x+y)^2 =x^2 + 2xy + y^2
If this statement is wrong, please give me a counterexample
cu
cx
peano-axioms
peano-axioms
asked Jan 29 at 15:27
user508589user508589
12
asked Jan 29 at 15:27
user508589user508589
12
asked Jan 29 at 15:27
user508589user508589
12
asked Jan 29 at 15:27
user508589user508589
12
asked Jan 29 at 15:27
user508589user508589
12
12
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
EDIT: It looks like I misread your question. There is an important difference between quantifier-free sentences, which is what my original answer (now below the fold) addressed, and quantifier-free formulas. Namely, when you speak of PA proving a formula with free variables (or of a structure satisfying such a formula), you're really asking about PA proving (or a structure satisfying) the universal closure of a quantifier-free formula: e.g. $$PAvdash (x+y)^2=x^2+2x+y^2$$ is really shorthand for $$PAvdashforall x,y((x+y)^2=x^2+2x+y^2).$$
Such statements are no longer provable in PA in general. For example, it's essentially a consequence of the MRDP theorem that there is a Diophantine equation $t_1(x_1,..,x_n)=t_2(x_1,...,x_n)$ which has no solutions but which PA can't prove has no solutions; then the quantifier-free formula $$neg(t_1(x_1,...,x_n)=t_2(x_1,...,x_n))$$ is true in $mathbb{N}$ but not PA-provable.
What if we don't work with implicit universal quantifiers - that is, we restrict attention to sentences (= formulas with no free variables)? This was the original - incorrect - way I interpreted your question.
Here we get a positive result: PA proves every true $Sigma_1$ sentence (that is, every sentence of the form $exists x_1...exists x_nvarphi(x_1,...,x_n)$, where $varphi$ uses only bounded quantifiers). This fact ("$Sigma_1$-completeness") about PA does not depend on PA being consistent - it is provable in PA itself, or indeed much less. The proof is tedious but not hard. Under the further assumption that PA is $Sigma_1$-sound, this implies that the true $Sigma_1$-sentences are exactly those which PA proves.
So this exactly delineates PA's power in terms of the coarse arithmetic hierarchy: PA proves exactly the correct $Sigma_1$ sentences (under a reasonable assumption on PA), but there are correct $Pi_1$ sentences which PA doesn't prove.
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
$endgroup$
$begingroup$
Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx
$endgroup$
– user508589
Jan 29 at 21:29
$begingroup$
@user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited.
$endgroup$
– Noah Schweber
Jan 29 at 22:35
add a comment |
$begingroup$
$$
text{You gave a counterexample: true in N but not PA-provable.} \
text{But there are so many examples with: true in N and PA-provable: } \
text{I define: } \
text{I_N} = {0,1,2, ...} text{ is the set of intuitive natural numbers} \
bar 0 ; and ; bar 1 text{ are the constants from PA} \
oplus and circ text{ are the functions in PA} \
text{when z is element from I_N, I define:} \
bar z := bar 1 oplus ... oplus + bar 1 quad text{n-times} \
text{I proofed the following statement "S1"} \
text{When n is element from I_N then we have:} \
PA vdash forall x forall y ; (bar z = x oplus y quad .rightarrow. quad
bigvee_{i=0}^z (x=bar i land y=overline{z-i}) \
text{I assume, that there are "many" of "similar" statements like S1.} \
text{Is it possible to formulate a statement "S", so that "S1" follows from "S" ?} \
$$
cu
cx
answered Feb 3 at 9:42
user508589user508589
12
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
EDIT: It looks like I misread your question. There is an important difference between quantifier-free sentences, which is what my original answer (now below the fold) addressed, and quantifier-free formulas. Namely, when you speak of PA proving a formula with free variables (or of a structure satisfying such a formula), you're really asking about PA proving (or a structure satisfying) the universal closure of a quantifier-free formula: e.g. $$PAvdash (x+y)^2=x^2+2x+y^2$$ is really shorthand for $$PAvdashforall x,y((x+y)^2=x^2+2x+y^2).$$
Such statements are no longer provable in PA in general. For example, it's essentially a consequence of the MRDP theorem that there is a Diophantine equation $t_1(x_1,..,x_n)=t_2(x_1,...,x_n)$ which has no solutions but which PA can't prove has no solutions; then the quantifier-free formula $$neg(t_1(x_1,...,x_n)=t_2(x_1,...,x_n))$$ is true in $mathbb{N}$ but not PA-provable.
What if we don't work with implicit universal quantifiers - that is, we restrict attention to sentences (= formulas with no free variables)? This was the original - incorrect - way I interpreted your question.
Here we get a positive result: PA proves every true $Sigma_1$ sentence (that is, every sentence of the form $exists x_1...exists x_nvarphi(x_1,...,x_n)$, where $varphi$ uses only bounded quantifiers). This fact ("$Sigma_1$-completeness") about PA does not depend on PA being consistent - it is provable in PA itself, or indeed much less. The proof is tedious but not hard. Under the further assumption that PA is $Sigma_1$-sound, this implies that the true $Sigma_1$-sentences are exactly those which PA proves.
So this exactly delineates PA's power in terms of the coarse arithmetic hierarchy: PA proves exactly the correct $Sigma_1$ sentences (under a reasonable assumption on PA), but there are correct $Pi_1$ sentences which PA doesn't prove.
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
$endgroup$
$begingroup$
Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx
$endgroup$
– user508589
Jan 29 at 21:29
$begingroup$
@user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited.
$endgroup$
– Noah Schweber
Jan 29 at 22:35
add a comment |
$begingroup$
EDIT: It looks like I misread your question. There is an important difference between quantifier-free sentences, which is what my original answer (now below the fold) addressed, and quantifier-free formulas. Namely, when you speak of PA proving a formula with free variables (or of a structure satisfying such a formula), you're really asking about PA proving (or a structure satisfying) the universal closure of a quantifier-free formula: e.g. $$PAvdash (x+y)^2=x^2+2x+y^2$$ is really shorthand for $$PAvdashforall x,y((x+y)^2=x^2+2x+y^2).$$
Such statements are no longer provable in PA in general. For example, it's essentially a consequence of the MRDP theorem that there is a Diophantine equation $t_1(x_1,..,x_n)=t_2(x_1,...,x_n)$ which has no solutions but which PA can't prove has no solutions; then the quantifier-free formula $$neg(t_1(x_1,...,x_n)=t_2(x_1,...,x_n))$$ is true in $mathbb{N}$ but not PA-provable.
What if we don't work with implicit universal quantifiers - that is, we restrict attention to sentences (= formulas with no free variables)? This was the original - incorrect - way I interpreted your question.
Here we get a positive result: PA proves every true $Sigma_1$ sentence (that is, every sentence of the form $exists x_1...exists x_nvarphi(x_1,...,x_n)$, where $varphi$ uses only bounded quantifiers). This fact ("$Sigma_1$-completeness") about PA does not depend on PA being consistent - it is provable in PA itself, or indeed much less. The proof is tedious but not hard. Under the further assumption that PA is $Sigma_1$-sound, this implies that the true $Sigma_1$-sentences are exactly those which PA proves.
So this exactly delineates PA's power in terms of the coarse arithmetic hierarchy: PA proves exactly the correct $Sigma_1$ sentences (under a reasonable assumption on PA), but there are correct $Pi_1$ sentences which PA doesn't prove.
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
$endgroup$
$begingroup$
Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx
$endgroup$
– user508589
Jan 29 at 21:29
$begingroup$
@user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited.
$endgroup$
– Noah Schweber
Jan 29 at 22:35
add a comment |
$begingroup$
EDIT: It looks like I misread your question. There is an important difference between quantifier-free sentences, which is what my original answer (now below the fold) addressed, and quantifier-free formulas. Namely, when you speak of PA proving a formula with free variables (or of a structure satisfying such a formula), you're really asking about PA proving (or a structure satisfying) the universal closure of a quantifier-free formula: e.g. $$PAvdash (x+y)^2=x^2+2x+y^2$$ is really shorthand for $$PAvdashforall x,y((x+y)^2=x^2+2x+y^2).$$
Such statements are no longer provable in PA in general. For example, it's essentially a consequence of the MRDP theorem that there is a Diophantine equation $t_1(x_1,..,x_n)=t_2(x_1,...,x_n)$ which has no solutions but which PA can't prove has no solutions; then the quantifier-free formula $$neg(t_1(x_1,...,x_n)=t_2(x_1,...,x_n))$$ is true in $mathbb{N}$ but not PA-provable.
What if we don't work with implicit universal quantifiers - that is, we restrict attention to sentences (= formulas with no free variables)? This was the original - incorrect - way I interpreted your question.
Here we get a positive result: PA proves every true $Sigma_1$ sentence (that is, every sentence of the form $exists x_1...exists x_nvarphi(x_1,...,x_n)$, where $varphi$ uses only bounded quantifiers). This fact ("$Sigma_1$-completeness") about PA does not depend on PA being consistent - it is provable in PA itself, or indeed much less. The proof is tedious but not hard. Under the further assumption that PA is $Sigma_1$-sound, this implies that the true $Sigma_1$-sentences are exactly those which PA proves.
So this exactly delineates PA's power in terms of the coarse arithmetic hierarchy: PA proves exactly the correct $Sigma_1$ sentences (under a reasonable assumption on PA), but there are correct $Pi_1$ sentences which PA doesn't prove.
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
$endgroup$
EDIT: It looks like I misread your question. There is an important difference between quantifier-free sentences, which is what my original answer (now below the fold) addressed, and quantifier-free formulas. Namely, when you speak of PA proving a formula with free variables (or of a structure satisfying such a formula), you're really asking about PA proving (or a structure satisfying) the universal closure of a quantifier-free formula: e.g. $$PAvdash (x+y)^2=x^2+2x+y^2$$ is really shorthand for $$PAvdashforall x,y((x+y)^2=x^2+2x+y^2).$$
Such statements are no longer provable in PA in general. For example, it's essentially a consequence of the MRDP theorem that there is a Diophantine equation $t_1(x_1,..,x_n)=t_2(x_1,...,x_n)$ which has no solutions but which PA can't prove has no solutions; then the quantifier-free formula $$neg(t_1(x_1,...,x_n)=t_2(x_1,...,x_n))$$ is true in $mathbb{N}$ but not PA-provable.
What if we don't work with implicit universal quantifiers - that is, we restrict attention to sentences (= formulas with no free variables)? This was the original - incorrect - way I interpreted your question.
Here we get a positive result: PA proves every true $Sigma_1$ sentence (that is, every sentence of the form $exists x_1...exists x_nvarphi(x_1,...,x_n)$, where $varphi$ uses only bounded quantifiers). This fact ("$Sigma_1$-completeness") about PA does not depend on PA being consistent - it is provable in PA itself, or indeed much less. The proof is tedious but not hard. Under the further assumption that PA is $Sigma_1$-sound, this implies that the true $Sigma_1$-sentences are exactly those which PA proves.
So this exactly delineates PA's power in terms of the coarse arithmetic hierarchy: PA proves exactly the correct $Sigma_1$ sentences (under a reasonable assumption on PA), but there are correct $Pi_1$ sentences which PA doesn't prove.
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
edited Jan 29 at 22:56
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
answered Jan 29 at 16:02
Noah SchweberNoah Schweber
128k10151294
128k10151294
$begingroup$
Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx
$endgroup$
– user508589
Jan 29 at 21:29
$begingroup$
@user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited.
$endgroup$
– Noah Schweber
Jan 29 at 22:35
add a comment |
$begingroup$
Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx
$endgroup$
– user508589
Jan 29 at 21:29
$begingroup$
@user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited.
$endgroup$
– Noah Schweber
Jan 29 at 22:35
$begingroup$
Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx
$endgroup$
– user508589
Jan 29 at 21:29
$begingroup$
Thank you for your feedback. I tried to proof my statement, but I could not do it. My idee: every term represent as a polynom. Please can you give me a link to the proof of my statement. cu cx
$endgroup$
– user508589
Jan 29 at 21:29
$begingroup$
@user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited.
$endgroup$
– Noah Schweber
Jan 29 at 22:35
$begingroup$
@user508589 Looking at your question again, I misread it (the issue being "formula" versus "sentence" and the implicit universal quantifier when we talk about proving/satisfying the former). I've edited.
$endgroup$
– Noah Schweber
Jan 29 at 22:35
add a comment |
$begingroup$
$$
text{You gave a counterexample: true in N but not PA-provable.} \
text{But there are so many examples with: true in N and PA-provable: } \
text{I define: } \
text{I_N} = {0,1,2, ...} text{ is the set of intuitive natural numbers} \
bar 0 ; and ; bar 1 text{ are the constants from PA} \
oplus and circ text{ are the functions in PA} \
text{when z is element from I_N, I define:} \
bar z := bar 1 oplus ... oplus + bar 1 quad text{n-times} \
text{I proofed the following statement "S1"} \
text{When n is element from I_N then we have:} \
PA vdash forall x forall y ; (bar z = x oplus y quad .rightarrow. quad
bigvee_{i=0}^z (x=bar i land y=overline{z-i}) \
text{I assume, that there are "many" of "similar" statements like S1.} \
text{Is it possible to formulate a statement "S", so that "S1" follows from "S" ?} \
$$
cu
cx
answered Feb 3 at 9:42
user508589user508589
12
$endgroup$
add a comment |
$begingroup$
$$
text{You gave a counterexample: true in N but not PA-provable.} \
text{But there are so many examples with: true in N and PA-provable: } \
text{I define: } \
text{I_N} = {0,1,2, ...} text{ is the set of intuitive natural numbers} \
bar 0 ; and ; bar 1 text{ are the constants from PA} \
oplus and circ text{ are the functions in PA} \
text{when z is element from I_N, I define:} \
bar z := bar 1 oplus ... oplus + bar 1 quad text{n-times} \
text{I proofed the following statement "S1"} \
text{When n is element from I_N then we have:} \
PA vdash forall x forall y ; (bar z = x oplus y quad .rightarrow. quad
bigvee_{i=0}^z (x=bar i land y=overline{z-i}) \
text{I assume, that there are "many" of "similar" statements like S1.} \
text{Is it possible to formulate a statement "S", so that "S1" follows from "S" ?} \
$$
cu
cx
answered Feb 3 at 9:42
user508589user508589
12
$endgroup$
add a comment |
$begingroup$
$$
text{You gave a counterexample: true in N but not PA-provable.} \
text{But there are so many examples with: true in N and PA-provable: } \
text{I define: } \
text{I_N} = {0,1,2, ...} text{ is the set of intuitive natural numbers} \
bar 0 ; and ; bar 1 text{ are the constants from PA} \
oplus and circ text{ are the functions in PA} \
text{when z is element from I_N, I define:} \
bar z := bar 1 oplus ... oplus + bar 1 quad text{n-times} \
text{I proofed the following statement "S1"} \
text{When n is element from I_N then we have:} \
PA vdash forall x forall y ; (bar z = x oplus y quad .rightarrow. quad
bigvee_{i=0}^z (x=bar i land y=overline{z-i}) \
text{I assume, that there are "many" of "similar" statements like S1.} \
text{Is it possible to formulate a statement "S", so that "S1" follows from "S" ?} \
$$
cu
cx
answered Feb 3 at 9:42
user508589user508589
12
$endgroup$
$$
text{You gave a counterexample: true in N but not PA-provable.} \
text{But there are so many examples with: true in N and PA-provable: } \
text{I define: } \
text{I_N} = {0,1,2, ...} text{ is the set of intuitive natural numbers} \
bar 0 ; and ; bar 1 text{ are the constants from PA} \
oplus and circ text{ are the functions in PA} \
text{when z is element from I_N, I define:} \
bar z := bar 1 oplus ... oplus + bar 1 quad text{n-times} \
text{I proofed the following statement "S1"} \
text{When n is element from I_N then we have:} \
PA vdash forall x forall y ; (bar z = x oplus y quad .rightarrow. quad
bigvee_{i=0}^z (x=bar i land y=overline{z-i}) \
text{I assume, that there are "many" of "similar" statements like S1.} \
text{Is it possible to formulate a statement "S", so that "S1" follows from "S" ?} \
$$
cu
cx
answered Feb 3 at 9:42
user508589user508589
12
edited Feb 4 at 9:09
answered Feb 3 at 9:42
user508589user508589
12
answered Feb 3 at 9:42
user508589user508589
12
answered Feb 3 at 9:42
user508589user508589
12
12
add a comment |
add a comment |
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