Difference between first and second order induction?

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Can anyone explain the difference between induction as it's stated in first order logic and that from second order logic? I don't understand the difference as it pertains to things like Peano axioms.
logic induction definition peano-axioms
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Can anyone explain the difference between induction as it's stated in first order logic and that from second order logic? I don't understand the difference as it pertains to things like Peano axioms.
logic induction definition peano-axioms
3
Second-order induction is an axiom which applies to any set $Xsubseteq mathbbN$: if $0in X$ and $forall n(n in X rightarrow n+1in X)$, then $X=mathbbN$. First-order induction is an axiom (or, to be rigorous, a scheme of axioms) which works the same way but only applies to those set defined by some first-order formula $phi$, i.e., sets of the form $X= nin mathbbN: phi(n) mathrmis true$.
â realdonaldtrump
3 hours ago
@realdonaldtrump: of course, when we work in second-order arithmetic, the "first order induction scheme" is typically defined to include the formula "$n in X$", and so the second-order induction axiom is just one particular axiom in the induction scheme.
â Carl Mummert
4 mins ago
@user525966 - it is not completely clear to me what you're asking. Which induction statement in second order logic are you looking at? What kind of second order theories are you interested in? The induction axioms aren't really part of logic, they are specific to theories of arithmetic.
â Carl Mummert
2 mins ago
add a comment |Â
up vote
2
down vote
favorite
up vote
2
down vote
favorite
Can anyone explain the difference between induction as it's stated in first order logic and that from second order logic? I don't understand the difference as it pertains to things like Peano axioms.
logic induction definition peano-axioms
Can anyone explain the difference between induction as it's stated in first order logic and that from second order logic? I don't understand the difference as it pertains to things like Peano axioms.
logic induction definition peano-axioms
logic induction definition peano-axioms
asked 3 hours ago
user525966
1,943821
1,943821
3
Second-order induction is an axiom which applies to any set $Xsubseteq mathbbN$: if $0in X$ and $forall n(n in X rightarrow n+1in X)$, then $X=mathbbN$. First-order induction is an axiom (or, to be rigorous, a scheme of axioms) which works the same way but only applies to those set defined by some first-order formula $phi$, i.e., sets of the form $X= nin mathbbN: phi(n) mathrmis true$.
â realdonaldtrump
3 hours ago
@realdonaldtrump: of course, when we work in second-order arithmetic, the "first order induction scheme" is typically defined to include the formula "$n in X$", and so the second-order induction axiom is just one particular axiom in the induction scheme.
â Carl Mummert
4 mins ago
@user525966 - it is not completely clear to me what you're asking. Which induction statement in second order logic are you looking at? What kind of second order theories are you interested in? The induction axioms aren't really part of logic, they are specific to theories of arithmetic.
â Carl Mummert
2 mins ago
add a comment |Â
3
Second-order induction is an axiom which applies to any set $Xsubseteq mathbbN$: if $0in X$ and $forall n(n in X rightarrow n+1in X)$, then $X=mathbbN$. First-order induction is an axiom (or, to be rigorous, a scheme of axioms) which works the same way but only applies to those set defined by some first-order formula $phi$, i.e., sets of the form $X= nin mathbbN: phi(n) mathrmis true$.
â realdonaldtrump
3 hours ago
@realdonaldtrump: of course, when we work in second-order arithmetic, the "first order induction scheme" is typically defined to include the formula "$n in X$", and so the second-order induction axiom is just one particular axiom in the induction scheme.
â Carl Mummert
4 mins ago
@user525966 - it is not completely clear to me what you're asking. Which induction statement in second order logic are you looking at? What kind of second order theories are you interested in? The induction axioms aren't really part of logic, they are specific to theories of arithmetic.
â Carl Mummert
2 mins ago
3
3
Second-order induction is an axiom which applies to any set $Xsubseteq mathbbN$: if $0in X$ and $forall n(n in X rightarrow n+1in X)$, then $X=mathbbN$. First-order induction is an axiom (or, to be rigorous, a scheme of axioms) which works the same way but only applies to those set defined by some first-order formula $phi$, i.e., sets of the form $X= nin mathbbN: phi(n) mathrmis true$.
â realdonaldtrump
3 hours ago
Second-order induction is an axiom which applies to any set $Xsubseteq mathbbN$: if $0in X$ and $forall n(n in X rightarrow n+1in X)$, then $X=mathbbN$. First-order induction is an axiom (or, to be rigorous, a scheme of axioms) which works the same way but only applies to those set defined by some first-order formula $phi$, i.e., sets of the form $X= nin mathbbN: phi(n) mathrmis true$.
â realdonaldtrump
3 hours ago
@realdonaldtrump: of course, when we work in second-order arithmetic, the "first order induction scheme" is typically defined to include the formula "$n in X$", and so the second-order induction axiom is just one particular axiom in the induction scheme.
â Carl Mummert
4 mins ago
@realdonaldtrump: of course, when we work in second-order arithmetic, the "first order induction scheme" is typically defined to include the formula "$n in X$", and so the second-order induction axiom is just one particular axiom in the induction scheme.
â Carl Mummert
4 mins ago
@user525966 - it is not completely clear to me what you're asking. Which induction statement in second order logic are you looking at? What kind of second order theories are you interested in? The induction axioms aren't really part of logic, they are specific to theories of arithmetic.
â Carl Mummert
2 mins ago
@user525966 - it is not completely clear to me what you're asking. Which induction statement in second order logic are you looking at? What kind of second order theories are you interested in? The induction axioms aren't really part of logic, they are specific to theories of arithmetic.
â Carl Mummert
2 mins ago
add a comment |Â
1 Answer
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The informal statement of induction is:
For any property $P$ of natural numbers, if $P(0)$ holds, and $P(n)$ implies $P(n+1)$ for all $n$, then $P(n)$ holds for all $n$.
Of course, this raises the question: What exactly do we mean by a "property of natural numbers"?
One natural interpretation is to identify properties of natural numbers with sets of natural numbers. That is, for any property $P$, we can form the set of all natural numbers satisfying that property. And for any set of natural numbers $X$, we can consider the property of being in $X$. For example, the property of being a prime number corresponds to the set $nin mathbbNmid ntext is prime$
Another natural interpretation is to identify properties of natural numbers with formulas in one free variable in some logic (in this discussion, let's just talk about first-order logic in the language of arithemetic). Here the syntax of the logic gives us a language for writing down properties of natural numbers. For example, the property of being a prime number corresponds to the formula $lnot (x= 1)land forall y, (exists z, (ycdot z = x) rightarrow (y = 1 lor y = x))$.
Induction under the interpretation "properties are sets" can be formalized as follows:
$forall Psubseteq mathbbN: ((0in Pland forall nin mathbbN: (nin P rightarrow (n+1)in P))rightarrow forall nin mathbbN: nin P)$
This is a sentence of second-order logic, since it involves a quantification $forall Psubseteq mathbbN$ over subsets of $mathbbN$.
The interpretation "properties are formulas" leads to the following formalization of induction:
$(varphi(0)land forall n, (varphi(n)rightarrow varphi(n+1)) rightarrow forall n,varphi(n)$
Here we have an infinite schema of sentences of first-order logic, one for each first-order formula $varphi(x)$. It's first-order because the quantifiers only range over elements of $mathbbN$, not subsets, and the formulas $varphi(x)$ are themselves first-order.
It's worth noting that second-order induction is much stronger than first-order induction. Second-order induction applies to all subsets, while first-order induction only applies to those which can be defined by some first-order formula (and since there are are $2^aleph_0$-many subsets of $mathbbN$ and only $aleph_0$-many first-order formulas, there are many subsets which are not definable).
The second-order Peano axioms (which consist of some basic rules of arithmetic, together with the second-order induction axiom above) suffice to pin down $mathbbN$ up to isomorphism.
The first-order Peano axioms (which consist of some basic rules of arithmetic, together with the first-order induction axiom schema above) cannot hope to pin down $mathbbN$ up to isomorphism (thanks to the Löwenheim-Skolem theorems). That is, there are "non-standard models" of the first-order Peano axioms, which satisfy induction for all first-order definable properties, but not for arbitrary subsets.
It is true that second-order induction is stronger than first-order, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using second-order induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs - essentially reducing things to the question of which formulas have a comprehension axiom.
â Carl Mummert
17 mins ago
Of course, when we work semantically with full second order semantics, in essence we have comprehension for all subsets of N, and so second-order induction is much stronger than the first-order induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex.
â Carl Mummert
13 mins ago
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
The informal statement of induction is:
For any property $P$ of natural numbers, if $P(0)$ holds, and $P(n)$ implies $P(n+1)$ for all $n$, then $P(n)$ holds for all $n$.
Of course, this raises the question: What exactly do we mean by a "property of natural numbers"?
One natural interpretation is to identify properties of natural numbers with sets of natural numbers. That is, for any property $P$, we can form the set of all natural numbers satisfying that property. And for any set of natural numbers $X$, we can consider the property of being in $X$. For example, the property of being a prime number corresponds to the set $nin mathbbNmid ntext is prime$
Another natural interpretation is to identify properties of natural numbers with formulas in one free variable in some logic (in this discussion, let's just talk about first-order logic in the language of arithemetic). Here the syntax of the logic gives us a language for writing down properties of natural numbers. For example, the property of being a prime number corresponds to the formula $lnot (x= 1)land forall y, (exists z, (ycdot z = x) rightarrow (y = 1 lor y = x))$.
Induction under the interpretation "properties are sets" can be formalized as follows:
$forall Psubseteq mathbbN: ((0in Pland forall nin mathbbN: (nin P rightarrow (n+1)in P))rightarrow forall nin mathbbN: nin P)$
This is a sentence of second-order logic, since it involves a quantification $forall Psubseteq mathbbN$ over subsets of $mathbbN$.
The interpretation "properties are formulas" leads to the following formalization of induction:
$(varphi(0)land forall n, (varphi(n)rightarrow varphi(n+1)) rightarrow forall n,varphi(n)$
Here we have an infinite schema of sentences of first-order logic, one for each first-order formula $varphi(x)$. It's first-order because the quantifiers only range over elements of $mathbbN$, not subsets, and the formulas $varphi(x)$ are themselves first-order.
It's worth noting that second-order induction is much stronger than first-order induction. Second-order induction applies to all subsets, while first-order induction only applies to those which can be defined by some first-order formula (and since there are are $2^aleph_0$-many subsets of $mathbbN$ and only $aleph_0$-many first-order formulas, there are many subsets which are not definable).
The second-order Peano axioms (which consist of some basic rules of arithmetic, together with the second-order induction axiom above) suffice to pin down $mathbbN$ up to isomorphism.
The first-order Peano axioms (which consist of some basic rules of arithmetic, together with the first-order induction axiom schema above) cannot hope to pin down $mathbbN$ up to isomorphism (thanks to the Löwenheim-Skolem theorems). That is, there are "non-standard models" of the first-order Peano axioms, which satisfy induction for all first-order definable properties, but not for arbitrary subsets.
It is true that second-order induction is stronger than first-order, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using second-order induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs - essentially reducing things to the question of which formulas have a comprehension axiom.
â Carl Mummert
17 mins ago
Of course, when we work semantically with full second order semantics, in essence we have comprehension for all subsets of N, and so second-order induction is much stronger than the first-order induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex.
â Carl Mummert
13 mins ago
add a comment |Â
up vote
4
down vote
The informal statement of induction is:
For any property $P$ of natural numbers, if $P(0)$ holds, and $P(n)$ implies $P(n+1)$ for all $n$, then $P(n)$ holds for all $n$.
Of course, this raises the question: What exactly do we mean by a "property of natural numbers"?
One natural interpretation is to identify properties of natural numbers with sets of natural numbers. That is, for any property $P$, we can form the set of all natural numbers satisfying that property. And for any set of natural numbers $X$, we can consider the property of being in $X$. For example, the property of being a prime number corresponds to the set $nin mathbbNmid ntext is prime$
Another natural interpretation is to identify properties of natural numbers with formulas in one free variable in some logic (in this discussion, let's just talk about first-order logic in the language of arithemetic). Here the syntax of the logic gives us a language for writing down properties of natural numbers. For example, the property of being a prime number corresponds to the formula $lnot (x= 1)land forall y, (exists z, (ycdot z = x) rightarrow (y = 1 lor y = x))$.
Induction under the interpretation "properties are sets" can be formalized as follows:
$forall Psubseteq mathbbN: ((0in Pland forall nin mathbbN: (nin P rightarrow (n+1)in P))rightarrow forall nin mathbbN: nin P)$
This is a sentence of second-order logic, since it involves a quantification $forall Psubseteq mathbbN$ over subsets of $mathbbN$.
The interpretation "properties are formulas" leads to the following formalization of induction:
$(varphi(0)land forall n, (varphi(n)rightarrow varphi(n+1)) rightarrow forall n,varphi(n)$
Here we have an infinite schema of sentences of first-order logic, one for each first-order formula $varphi(x)$. It's first-order because the quantifiers only range over elements of $mathbbN$, not subsets, and the formulas $varphi(x)$ are themselves first-order.
It's worth noting that second-order induction is much stronger than first-order induction. Second-order induction applies to all subsets, while first-order induction only applies to those which can be defined by some first-order formula (and since there are are $2^aleph_0$-many subsets of $mathbbN$ and only $aleph_0$-many first-order formulas, there are many subsets which are not definable).
The second-order Peano axioms (which consist of some basic rules of arithmetic, together with the second-order induction axiom above) suffice to pin down $mathbbN$ up to isomorphism.
The first-order Peano axioms (which consist of some basic rules of arithmetic, together with the first-order induction axiom schema above) cannot hope to pin down $mathbbN$ up to isomorphism (thanks to the Löwenheim-Skolem theorems). That is, there are "non-standard models" of the first-order Peano axioms, which satisfy induction for all first-order definable properties, but not for arbitrary subsets.
It is true that second-order induction is stronger than first-order, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using second-order induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs - essentially reducing things to the question of which formulas have a comprehension axiom.
â Carl Mummert
17 mins ago
Of course, when we work semantically with full second order semantics, in essence we have comprehension for all subsets of N, and so second-order induction is much stronger than the first-order induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex.
â Carl Mummert
13 mins ago
add a comment |Â
up vote
4
down vote
up vote
4
down vote
The informal statement of induction is:
For any property $P$ of natural numbers, if $P(0)$ holds, and $P(n)$ implies $P(n+1)$ for all $n$, then $P(n)$ holds for all $n$.
Of course, this raises the question: What exactly do we mean by a "property of natural numbers"?
One natural interpretation is to identify properties of natural numbers with sets of natural numbers. That is, for any property $P$, we can form the set of all natural numbers satisfying that property. And for any set of natural numbers $X$, we can consider the property of being in $X$. For example, the property of being a prime number corresponds to the set $nin mathbbNmid ntext is prime$
Another natural interpretation is to identify properties of natural numbers with formulas in one free variable in some logic (in this discussion, let's just talk about first-order logic in the language of arithemetic). Here the syntax of the logic gives us a language for writing down properties of natural numbers. For example, the property of being a prime number corresponds to the formula $lnot (x= 1)land forall y, (exists z, (ycdot z = x) rightarrow (y = 1 lor y = x))$.
Induction under the interpretation "properties are sets" can be formalized as follows:
$forall Psubseteq mathbbN: ((0in Pland forall nin mathbbN: (nin P rightarrow (n+1)in P))rightarrow forall nin mathbbN: nin P)$
This is a sentence of second-order logic, since it involves a quantification $forall Psubseteq mathbbN$ over subsets of $mathbbN$.
The interpretation "properties are formulas" leads to the following formalization of induction:
$(varphi(0)land forall n, (varphi(n)rightarrow varphi(n+1)) rightarrow forall n,varphi(n)$
Here we have an infinite schema of sentences of first-order logic, one for each first-order formula $varphi(x)$. It's first-order because the quantifiers only range over elements of $mathbbN$, not subsets, and the formulas $varphi(x)$ are themselves first-order.
It's worth noting that second-order induction is much stronger than first-order induction. Second-order induction applies to all subsets, while first-order induction only applies to those which can be defined by some first-order formula (and since there are are $2^aleph_0$-many subsets of $mathbbN$ and only $aleph_0$-many first-order formulas, there are many subsets which are not definable).
The second-order Peano axioms (which consist of some basic rules of arithmetic, together with the second-order induction axiom above) suffice to pin down $mathbbN$ up to isomorphism.
The first-order Peano axioms (which consist of some basic rules of arithmetic, together with the first-order induction axiom schema above) cannot hope to pin down $mathbbN$ up to isomorphism (thanks to the Löwenheim-Skolem theorems). That is, there are "non-standard models" of the first-order Peano axioms, which satisfy induction for all first-order definable properties, but not for arbitrary subsets.
The informal statement of induction is:
For any property $P$ of natural numbers, if $P(0)$ holds, and $P(n)$ implies $P(n+1)$ for all $n$, then $P(n)$ holds for all $n$.
Of course, this raises the question: What exactly do we mean by a "property of natural numbers"?
One natural interpretation is to identify properties of natural numbers with sets of natural numbers. That is, for any property $P$, we can form the set of all natural numbers satisfying that property. And for any set of natural numbers $X$, we can consider the property of being in $X$. For example, the property of being a prime number corresponds to the set $nin mathbbNmid ntext is prime$
Another natural interpretation is to identify properties of natural numbers with formulas in one free variable in some logic (in this discussion, let's just talk about first-order logic in the language of arithemetic). Here the syntax of the logic gives us a language for writing down properties of natural numbers. For example, the property of being a prime number corresponds to the formula $lnot (x= 1)land forall y, (exists z, (ycdot z = x) rightarrow (y = 1 lor y = x))$.
Induction under the interpretation "properties are sets" can be formalized as follows:
$forall Psubseteq mathbbN: ((0in Pland forall nin mathbbN: (nin P rightarrow (n+1)in P))rightarrow forall nin mathbbN: nin P)$
This is a sentence of second-order logic, since it involves a quantification $forall Psubseteq mathbbN$ over subsets of $mathbbN$.
The interpretation "properties are formulas" leads to the following formalization of induction:
$(varphi(0)land forall n, (varphi(n)rightarrow varphi(n+1)) rightarrow forall n,varphi(n)$
Here we have an infinite schema of sentences of first-order logic, one for each first-order formula $varphi(x)$. It's first-order because the quantifiers only range over elements of $mathbbN$, not subsets, and the formulas $varphi(x)$ are themselves first-order.
It's worth noting that second-order induction is much stronger than first-order induction. Second-order induction applies to all subsets, while first-order induction only applies to those which can be defined by some first-order formula (and since there are are $2^aleph_0$-many subsets of $mathbbN$ and only $aleph_0$-many first-order formulas, there are many subsets which are not definable).
The second-order Peano axioms (which consist of some basic rules of arithmetic, together with the second-order induction axiom above) suffice to pin down $mathbbN$ up to isomorphism.
The first-order Peano axioms (which consist of some basic rules of arithmetic, together with the first-order induction axiom schema above) cannot hope to pin down $mathbbN$ up to isomorphism (thanks to the Löwenheim-Skolem theorems). That is, there are "non-standard models" of the first-order Peano axioms, which satisfy induction for all first-order definable properties, but not for arbitrary subsets.
edited 2 hours ago
answered 2 hours ago
Alex Kruckman
25.3k22455
25.3k22455
It is true that second-order induction is stronger than first-order, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using second-order induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs - essentially reducing things to the question of which formulas have a comprehension axiom.
â Carl Mummert
17 mins ago
Of course, when we work semantically with full second order semantics, in essence we have comprehension for all subsets of N, and so second-order induction is much stronger than the first-order induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex.
â Carl Mummert
13 mins ago
add a comment |Â
It is true that second-order induction is stronger than first-order, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using second-order induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs - essentially reducing things to the question of which formulas have a comprehension axiom.
â Carl Mummert
17 mins ago
Of course, when we work semantically with full second order semantics, in essence we have comprehension for all subsets of N, and so second-order induction is much stronger than the first-order induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex.
â Carl Mummert
13 mins ago
It is true that second-order induction is stronger than first-order, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using second-order induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs - essentially reducing things to the question of which formulas have a comprehension axiom.
â Carl Mummert
17 mins ago
It is true that second-order induction is stronger than first-order, but only when we are working in a context where we have strong comprehension axioms as well. For example, if we are working syntactically, and we want to prove some formula using second-order induction, we will have to construct the set $P$ in our formal proof before we can apply induction to it, and so the strength of the induction axiom will depend on which sets we can construct in our proofs - essentially reducing things to the question of which formulas have a comprehension axiom.
â Carl Mummert
17 mins ago
Of course, when we work semantically with full second order semantics, in essence we have comprehension for all subsets of N, and so second-order induction is much stronger than the first-order induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex.
â Carl Mummert
13 mins ago
Of course, when we work semantically with full second order semantics, in essence we have comprehension for all subsets of N, and so second-order induction is much stronger than the first-order induction scheme in that setting. But when we begin to look at particular formal theories, the relationship is more complex.
â Carl Mummert
13 mins ago
add a comment |Â
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3
Second-order induction is an axiom which applies to any set $Xsubseteq mathbbN$: if $0in X$ and $forall n(n in X rightarrow n+1in X)$, then $X=mathbbN$. First-order induction is an axiom (or, to be rigorous, a scheme of axioms) which works the same way but only applies to those set defined by some first-order formula $phi$, i.e., sets of the form $X= nin mathbbN: phi(n) mathrmis true$.
â realdonaldtrump
3 hours ago
@realdonaldtrump: of course, when we work in second-order arithmetic, the "first order induction scheme" is typically defined to include the formula "$n in X$", and so the second-order induction axiom is just one particular axiom in the induction scheme.
â Carl Mummert
4 mins ago
@user525966 - it is not completely clear to me what you're asking. Which induction statement in second order logic are you looking at? What kind of second order theories are you interested in? The induction axioms aren't really part of logic, they are specific to theories of arithmetic.
â Carl Mummert
2 mins ago