How to define substitution using ZFC
$begingroup$
One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.
Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
$$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$
Any ideas?
substitution foundations
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
$endgroup$
add a comment |
$begingroup$
One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.
Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
$$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$
Any ideas?
substitution foundations
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
$endgroup$
add a comment |
$begingroup$
One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.
Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
$$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$
Any ideas?
substitution foundations
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
$endgroup$
One question I've had regarding ZFC is how to define substitution. I cannot see how it's possible, despite the frequent use of substitution within both pure and applied mathematics.
Just to be clear, I define substitution as follows: for all well-defined mathematical objects $a$ and $b$ that are equal, and any function $f$ such that $f(a)$ and $f(b)$ are well-defined, $f(a) = f(b)$.
$$ forall a forall b forall f [(a = b , wedge f(a),f(b) text{ are defined}) Rightarrow f(a) = f(b)]$$
Any ideas?
substitution foundations
substitution foundations
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
asked Dec 31 '18 at 12:22
HarrisonOHarrisonO
707
707
add a comment |
add a comment |
2 Answers
2
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oldest
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$begingroup$
Substitution is "governed" by the axioms for equality.
We can substitute equals into a formula $varphi$ :
$a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,
or we can substitute into a function $f$ :
$a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.
Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :
$a = b to ((z in a) to (z in b))$.
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
$endgroup$
$begingroup$
So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
$endgroup$
– HarrisonO
Dec 31 '18 at 23:50
1
$begingroup$
@HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:16
add a comment |
$begingroup$
You can actually formulate ZFC in FOL without equality: with extensionality reading
$ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $
define a binary predicate to mean the consequent in the last sentence
$ xRy :iff forall z ( x in z iff y in z ) $
reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair
$ forall x exists yforall z(z in y iff z in x wedge varphi) $
$ forall x forall y exists z (x in z wedge y in z) $
prove substitution for $ R $, ie:
$ xRy implies (varphi x iff varphi y) $
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
$endgroup$
add a comment |
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2 Answers
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2 Answers
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$begingroup$
Substitution is "governed" by the axioms for equality.
We can substitute equals into a formula $varphi$ :
$a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,
or we can substitute into a function $f$ :
$a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.
Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :
$a = b to ((z in a) to (z in b))$.
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
$endgroup$
$begingroup$
So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
$endgroup$
– HarrisonO
Dec 31 '18 at 23:50
1
$begingroup$
@HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:16
add a comment |
$begingroup$
Substitution is "governed" by the axioms for equality.
We can substitute equals into a formula $varphi$ :
$a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,
or we can substitute into a function $f$ :
$a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.
Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :
$a = b to ((z in a) to (z in b))$.
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
$endgroup$
$begingroup$
So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
$endgroup$
– HarrisonO
Dec 31 '18 at 23:50
1
$begingroup$
@HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:16
add a comment |
$begingroup$
Substitution is "governed" by the axioms for equality.
We can substitute equals into a formula $varphi$ :
$a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,
or we can substitute into a function $f$ :
$a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.
Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :
$a = b to ((z in a) to (z in b))$.
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
$endgroup$
Substitution is "governed" by the axioms for equality.
We can substitute equals into a formula $varphi$ :
$a = b → (φ[a/x] → φ[b/x])$, for any formula $φ(x)$,
or we can substitute into a function $f$ :
$a=b to f(ldots,a,ldots) = f(ldots,b,ldots)$, for any function symbol $f$.
Example in $mathsf {ZF}$, using the formula $(z in x)$ as $φ(x)$ :
$a = b to ((z in a) to (z in b))$.
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
answered Dec 31 '18 at 12:42
Mauro ALLEGRANZAMauro ALLEGRANZA
68.3k449118
68.3k449118
$begingroup$
So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
$endgroup$
– HarrisonO
Dec 31 '18 at 23:50
1
$begingroup$
@HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:16
add a comment |
$begingroup$
So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
$endgroup$
– HarrisonO
Dec 31 '18 at 23:50
1
$begingroup$
@HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:16
$begingroup$
So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
$endgroup$
– HarrisonO
Dec 31 '18 at 23:50
$begingroup$
So is it not possible to define substitution using the ZF or ZFC axioms, requiring the axioms for equality? Otherwise, can you please tell me the exact axioms needed. Thanks.
$endgroup$
– HarrisonO
Dec 31 '18 at 23:50
1
1
$begingroup$
@HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:16
$begingroup$
@HarrisonO - $mathsf {ZF}$ and $mathsf {ZFC}$ are "axiom systems formulated in first-order logic with equality and with only one binary relation symbol $∈$ for membership
$endgroup$
– Mauro ALLEGRANZA
Jan 2 at 10:16
add a comment |
$begingroup$
You can actually formulate ZFC in FOL without equality: with extensionality reading
$ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $
define a binary predicate to mean the consequent in the last sentence
$ xRy :iff forall z ( x in z iff y in z ) $
reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair
$ forall x exists yforall z(z in y iff z in x wedge varphi) $
$ forall x forall y exists z (x in z wedge y in z) $
prove substitution for $ R $, ie:
$ xRy implies (varphi x iff varphi y) $
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
$endgroup$
add a comment |
$begingroup$
You can actually formulate ZFC in FOL without equality: with extensionality reading
$ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $
define a binary predicate to mean the consequent in the last sentence
$ xRy :iff forall z ( x in z iff y in z ) $
reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair
$ forall x exists yforall z(z in y iff z in x wedge varphi) $
$ forall x forall y exists z (x in z wedge y in z) $
prove substitution for $ R $, ie:
$ xRy implies (varphi x iff varphi y) $
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
$endgroup$
add a comment |
$begingroup$
You can actually formulate ZFC in FOL without equality: with extensionality reading
$ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $
define a binary predicate to mean the consequent in the last sentence
$ xRy :iff forall z ( x in z iff y in z ) $
reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair
$ forall x exists yforall z(z in y iff z in x wedge varphi) $
$ forall x forall y exists z (x in z wedge y in z) $
prove substitution for $ R $, ie:
$ xRy implies (varphi x iff varphi y) $
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
$endgroup$
You can actually formulate ZFC in FOL without equality: with extensionality reading
$ forall z ( z in x iff z in y ) implies forall z ( x in z iff y in z ) $
define a binary predicate to mean the consequent in the last sentence
$ xRy :iff forall z ( x in z iff y in z ) $
reflexivity, transitivity and symmetry following easily from FOL alone; and then with the usual comprehension/separation scheme and weak-pair
$ forall x exists yforall z(z in y iff z in x wedge varphi) $
$ forall x forall y exists z (x in z wedge y in z) $
prove substitution for $ R $, ie:
$ xRy implies (varphi x iff varphi y) $
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
answered Feb 15 at 1:38
a.c.brunoa.c.bruno
48639
48639
add a comment |
add a comment |
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