Why is the “axiom of extension” an axiom? [duplicate]
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This question already has an answer here:
What's the point of the axiom of extensionality?
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I think this is a definition of = because it explains the meaning of the new symbol =. However, I wonder why the set theory thinks this as an axiom rather than a definition.
logic set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
asked Jan 21 at 8:52
amoogaeamoogae
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Jan 21 at 9:41
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$begingroup$
This question already has an answer here:
What's the point of the axiom of extensionality?
2 answers
I think this is a definition of = because it explains the meaning of the new symbol =. However, I wonder why the set theory thinks this as an axiom rather than a definition.
logic set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
asked Jan 21 at 8:52
amoogaeamoogae
487
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marked as duplicate by José Carlos Santos, Asaf Karagila♦ logic
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Jan 21 at 9:41
This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.
add a comment |
$begingroup$
This question already has an answer here:
What's the point of the axiom of extensionality?
2 answers
I think this is a definition of = because it explains the meaning of the new symbol =. However, I wonder why the set theory thinks this as an axiom rather than a definition.
logic set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
asked Jan 21 at 8:52
amoogaeamoogae
487
$endgroup$
This question already has an answer here:
What's the point of the axiom of extensionality?
2 answers
I think this is a definition of = because it explains the meaning of the new symbol =. However, I wonder why the set theory thinks this as an axiom rather than a definition.
This question already has an answer here:
What's the point of the axiom of extensionality?
2 answers
logic set-theory axioms
logic set-theory axioms
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
asked Jan 21 at 8:52
amoogaeamoogae
487
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
asked Jan 21 at 8:52
amoogaeamoogae
487
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
edited Jan 21 at 10:57
Mauro ALLEGRANZA
66.4k449115
66.4k449115
asked Jan 21 at 8:52
amoogaeamoogae
487
asked Jan 21 at 8:52
amoogaeamoogae
487
asked Jan 21 at 8:52
amoogaeamoogae
487
487
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Jan 21 at 9:41
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1 Answer
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$begingroup$
See Axiom of extensionality.
There are two possibilities :
(i) the underlying logic is predicate calculus with equality.
In this case, the symbol $=$ is already defined by the equality axiom and Extensionality is needed only to legislate the "interaction" with $in$ :
$∀x∀y∀z [(z ∈ x leftrightarrow z ∈ y) to x = y]$.
(ii) the underlying logic is predicate calculus without equality.
In this case you are right : we need a specific definition for equality :
$a=b =_{def} forall x [x in a leftrightarrow x in b]$
and a different version of Extensionality :
$forall z forall x forall y [x = y to (x in z to y in z)]$.
From them we can derive the usual properties of equality.
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
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$begingroup$
+1 Very enlightening.
$endgroup$
– drhab
Jan 21 at 9:22
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Thank you very much! Can you tell me exactly what the (slightly different) version of Extensionality is?
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– amoogae
Jan 21 at 9:36
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I just read your edited answer. Thank you so much!
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– amoogae
Jan 21 at 11:01
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1 Answer
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1 Answer
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active
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$begingroup$
See Axiom of extensionality.
There are two possibilities :
(i) the underlying logic is predicate calculus with equality.
In this case, the symbol $=$ is already defined by the equality axiom and Extensionality is needed only to legislate the "interaction" with $in$ :
$∀x∀y∀z [(z ∈ x leftrightarrow z ∈ y) to x = y]$.
(ii) the underlying logic is predicate calculus without equality.
In this case you are right : we need a specific definition for equality :
$a=b =_{def} forall x [x in a leftrightarrow x in b]$
and a different version of Extensionality :
$forall z forall x forall y [x = y to (x in z to y in z)]$.
From them we can derive the usual properties of equality.
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
$endgroup$
$begingroup$
+1 Very enlightening.
$endgroup$
– drhab
Jan 21 at 9:22
$begingroup$
Thank you very much! Can you tell me exactly what the (slightly different) version of Extensionality is?
$endgroup$
– amoogae
Jan 21 at 9:36
$begingroup$
I just read your edited answer. Thank you so much!
$endgroup$
– amoogae
Jan 21 at 11:01
add a comment |
$begingroup$
See Axiom of extensionality.
There are two possibilities :
(i) the underlying logic is predicate calculus with equality.
In this case, the symbol $=$ is already defined by the equality axiom and Extensionality is needed only to legislate the "interaction" with $in$ :
$∀x∀y∀z [(z ∈ x leftrightarrow z ∈ y) to x = y]$.
(ii) the underlying logic is predicate calculus without equality.
In this case you are right : we need a specific definition for equality :
$a=b =_{def} forall x [x in a leftrightarrow x in b]$
and a different version of Extensionality :
$forall z forall x forall y [x = y to (x in z to y in z)]$.
From them we can derive the usual properties of equality.
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
$endgroup$
$begingroup$
+1 Very enlightening.
$endgroup$
– drhab
Jan 21 at 9:22
$begingroup$
Thank you very much! Can you tell me exactly what the (slightly different) version of Extensionality is?
$endgroup$
– amoogae
Jan 21 at 9:36
$begingroup$
I just read your edited answer. Thank you so much!
$endgroup$
– amoogae
Jan 21 at 11:01
add a comment |
$begingroup$
See Axiom of extensionality.
There are two possibilities :
(i) the underlying logic is predicate calculus with equality.
In this case, the symbol $=$ is already defined by the equality axiom and Extensionality is needed only to legislate the "interaction" with $in$ :
$∀x∀y∀z [(z ∈ x leftrightarrow z ∈ y) to x = y]$.
(ii) the underlying logic is predicate calculus without equality.
In this case you are right : we need a specific definition for equality :
$a=b =_{def} forall x [x in a leftrightarrow x in b]$
and a different version of Extensionality :
$forall z forall x forall y [x = y to (x in z to y in z)]$.
From them we can derive the usual properties of equality.
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
$endgroup$
See Axiom of extensionality.
There are two possibilities :
(i) the underlying logic is predicate calculus with equality.
In this case, the symbol $=$ is already defined by the equality axiom and Extensionality is needed only to legislate the "interaction" with $in$ :
$∀x∀y∀z [(z ∈ x leftrightarrow z ∈ y) to x = y]$.
(ii) the underlying logic is predicate calculus without equality.
In this case you are right : we need a specific definition for equality :
$a=b =_{def} forall x [x in a leftrightarrow x in b]$
and a different version of Extensionality :
$forall z forall x forall y [x = y to (x in z to y in z)]$.
From them we can derive the usual properties of equality.
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
edited Jan 21 at 15:11
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
answered Jan 21 at 9:00
Mauro ALLEGRANZAMauro ALLEGRANZA
66.4k449115
66.4k449115
$begingroup$
+1 Very enlightening.
$endgroup$
– drhab
Jan 21 at 9:22
$begingroup$
Thank you very much! Can you tell me exactly what the (slightly different) version of Extensionality is?
$endgroup$
– amoogae
Jan 21 at 9:36
$begingroup$
I just read your edited answer. Thank you so much!
$endgroup$
– amoogae
Jan 21 at 11:01
add a comment |
$begingroup$
+1 Very enlightening.
$endgroup$
– drhab
Jan 21 at 9:22
$begingroup$
Thank you very much! Can you tell me exactly what the (slightly different) version of Extensionality is?
$endgroup$
– amoogae
Jan 21 at 9:36
$begingroup$
I just read your edited answer. Thank you so much!
$endgroup$
– amoogae
Jan 21 at 11:01
$begingroup$
+1 Very enlightening.
$endgroup$
– drhab
Jan 21 at 9:22
$begingroup$
+1 Very enlightening.
$endgroup$
– drhab
Jan 21 at 9:22
$begingroup$
Thank you very much! Can you tell me exactly what the (slightly different) version of Extensionality is?
$endgroup$
– amoogae
Jan 21 at 9:36
$begingroup$
Thank you very much! Can you tell me exactly what the (slightly different) version of Extensionality is?
$endgroup$
– amoogae
Jan 21 at 9:36
$begingroup$
I just read your edited answer. Thank you so much!
$endgroup$
– amoogae
Jan 21 at 11:01
$begingroup$
I just read your edited answer. Thank you so much!
$endgroup$
– amoogae
Jan 21 at 11:01
add a comment |
