Mixing
Is $a=bRightarrow a+c=b+c$ always true? [closed]
$begingroup$
Let $(A,+)$ be a Magma that means it is not a $field$ is the above identity still true?
I have encountered the Substitution law at my first proof that if $A$ is a Group then ist Right neutrall element is also left neutral. So it has to be an Axiom, I couldn't find the Axiom. Is there a set of These Basic Axioms. I have heard About the ZF Axioms but they just deal with sets and not with operations defined on the set
linear-algebra
asked Jan 15 at 8:14
RM777RM777
39112
$endgroup$
closed as off-topic by user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos Feb 6 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Let $(A,+)$ be a Magma that means it is not a $field$ is the above identity still true?
I have encountered the Substitution law at my first proof that if $A$ is a Group then ist Right neutrall element is also left neutral. So it has to be an Axiom, I couldn't find the Axiom. Is there a set of These Basic Axioms. I have heard About the ZF Axioms but they just deal with sets and not with operations defined on the set
linear-algebra
asked Jan 15 at 8:14
RM777RM777
39112
$endgroup$
closed as off-topic by user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos Feb 6 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I think this is an axiom in first-order logic.
$endgroup$
– YuiTo Cheng
Jan 15 at 8:24
2
$begingroup$
It is an instance of substitution axiom for equality ; see also this post
$endgroup$
– Mauro ALLEGRANZA
Jan 15 at 8:32
2
$begingroup$
The operation is a function $+ :Atimes A to A$. If $a=b$, then $(a,c)=(b,c)$, and therefore $ + (a,c)=+ (b,c)$
$endgroup$
– leibnewtz
Jan 15 at 9:02
1
$begingroup$
Since you don't want to accept my (correct) answer, and some unknown joker downvoted it out of spite, I'm deleting it. Note that Michael Rozenberg does not understand basic first-order logic, much less the first-order theory of groups, and his answer is simply wrong but he refuses to admit it.
$endgroup$
– user21820
Feb 5 at 17:12
2
$begingroup$
@user21820 That's not a valid reason to delete an answer. Please refrain from such behaviour in the future. Your question is useful to others, and we do not encourage self-vandalism. I have upvoted your answer as a sign of good faith. Regards,
$endgroup$
– Pedro Tamaroff♦
Feb 12 at 1:04
add a comment |
$begingroup$
Let $(A,+)$ be a Magma that means it is not a $field$ is the above identity still true?
I have encountered the Substitution law at my first proof that if $A$ is a Group then ist Right neutrall element is also left neutral. So it has to be an Axiom, I couldn't find the Axiom. Is there a set of These Basic Axioms. I have heard About the ZF Axioms but they just deal with sets and not with operations defined on the set
linear-algebra
asked Jan 15 at 8:14
RM777RM777
39112
$endgroup$
Let $(A,+)$ be a Magma that means it is not a $field$ is the above identity still true?
I have encountered the Substitution law at my first proof that if $A$ is a Group then ist Right neutrall element is also left neutral. So it has to be an Axiom, I couldn't find the Axiom. Is there a set of These Basic Axioms. I have heard About the ZF Axioms but they just deal with sets and not with operations defined on the set
linear-algebra
linear-algebra
asked Jan 15 at 8:14
RM777RM777
39112
asked Jan 15 at 8:14
RM777RM777
39112
asked Jan 15 at 8:14
RM777RM777
39112
asked Jan 15 at 8:14
RM777RM777
39112
asked Jan 15 at 8:14
RM777RM777
39112
39112
closed as off-topic by user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos Feb 6 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos Feb 6 at 8:45
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – user21820, Andrew, Cesareo, ancientmathematician, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
$begingroup$
I think this is an axiom in first-order logic.
$endgroup$
– YuiTo Cheng
Jan 15 at 8:24
2
$begingroup$
It is an instance of substitution axiom for equality ; see also this post
$endgroup$
– Mauro ALLEGRANZA
Jan 15 at 8:32
2
$begingroup$
The operation is a function $+ :Atimes A to A$. If $a=b$, then $(a,c)=(b,c)$, and therefore $ + (a,c)=+ (b,c)$
$endgroup$
– leibnewtz
Jan 15 at 9:02
1
$begingroup$
Since you don't want to accept my (correct) answer, and some unknown joker downvoted it out of spite, I'm deleting it. Note that Michael Rozenberg does not understand basic first-order logic, much less the first-order theory of groups, and his answer is simply wrong but he refuses to admit it.
$endgroup$
– user21820
Feb 5 at 17:12
2
$begingroup$
@user21820 That's not a valid reason to delete an answer. Please refrain from such behaviour in the future. Your question is useful to others, and we do not encourage self-vandalism. I have upvoted your answer as a sign of good faith. Regards,
$endgroup$
– Pedro Tamaroff♦
Feb 12 at 1:04
add a comment |
$begingroup$
I think this is an axiom in first-order logic.
$endgroup$
– YuiTo Cheng
Jan 15 at 8:24
2
$begingroup$
It is an instance of substitution axiom for equality ; see also this post
$endgroup$
– Mauro ALLEGRANZA
Jan 15 at 8:32
2
$begingroup$
The operation is a function $+ :Atimes A to A$. If $a=b$, then $(a,c)=(b,c)$, and therefore $ + (a,c)=+ (b,c)$
$endgroup$
– leibnewtz
Jan 15 at 9:02
1
$begingroup$
Since you don't want to accept my (correct) answer, and some unknown joker downvoted it out of spite, I'm deleting it. Note that Michael Rozenberg does not understand basic first-order logic, much less the first-order theory of groups, and his answer is simply wrong but he refuses to admit it.
$endgroup$
– user21820
Feb 5 at 17:12
2
$begingroup$
@user21820 That's not a valid reason to delete an answer. Please refrain from such behaviour in the future. Your question is useful to others, and we do not encourage self-vandalism. I have upvoted your answer as a sign of good faith. Regards,
$endgroup$
– Pedro Tamaroff♦
Feb 12 at 1:04
$begingroup$
I think this is an axiom in first-order logic.
$endgroup$
– YuiTo Cheng
Jan 15 at 8:24
$begingroup$
I think this is an axiom in first-order logic.
$endgroup$
– YuiTo Cheng
Jan 15 at 8:24
2
2
$begingroup$
It is an instance of substitution axiom for equality ; see also this post
$endgroup$
– Mauro ALLEGRANZA
Jan 15 at 8:32
$begingroup$
It is an instance of substitution axiom for equality ; see also this post
$endgroup$
– Mauro ALLEGRANZA
Jan 15 at 8:32
2
2
$begingroup$
The operation is a function $+ :Atimes A to A$. If $a=b$, then $(a,c)=(b,c)$, and therefore $ + (a,c)=+ (b,c)$
$endgroup$
– leibnewtz
Jan 15 at 9:02
$begingroup$
The operation is a function $+ :Atimes A to A$. If $a=b$, then $(a,c)=(b,c)$, and therefore $ + (a,c)=+ (b,c)$
$endgroup$
– leibnewtz
Jan 15 at 9:02
1
1
$begingroup$
Since you don't want to accept my (correct) answer, and some unknown joker downvoted it out of spite, I'm deleting it. Note that Michael Rozenberg does not understand basic first-order logic, much less the first-order theory of groups, and his answer is simply wrong but he refuses to admit it.
$endgroup$
– user21820
Feb 5 at 17:12
$begingroup$
Since you don't want to accept my (correct) answer, and some unknown joker downvoted it out of spite, I'm deleting it. Note that Michael Rozenberg does not understand basic first-order logic, much less the first-order theory of groups, and his answer is simply wrong but he refuses to admit it.
$endgroup$
– user21820
Feb 5 at 17:12
2
2
$begingroup$
@user21820 That's not a valid reason to delete an answer. Please refrain from such behaviour in the future. Your question is useful to others, and we do not encourage self-vandalism. I have upvoted your answer as a sign of good faith. Regards,
$endgroup$
– Pedro Tamaroff♦
Feb 12 at 1:04
$begingroup$
@user21820 That's not a valid reason to delete an answer. Please refrain from such behaviour in the future. Your question is useful to others, and we do not encourage self-vandalism. I have upvoted your answer as a sign of good faith. Regards,
$endgroup$
– Pedro Tamaroff♦
Feb 12 at 1:04
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Michael's answer is simply wrong. In the axiomatization of a magma $(A,+)$, the binary function-symbol $+$ is not a function but merely a symbol. Also, there are 2 main kinds of deductive-systems for first-order logic.
In a Hilbert-style system, there is only one inference rule, modus ponens, and everything else is captured by axioms. The logical axioms include axioms for equality, as Mauro pointed out.
In a natural-deduction (Gentzen-style or sequent-style or Fitch-style) system, equality is typically not captured by axioms but by inference rules.
In both kinds, equality is built into the deductive system because it is considered part of first-order logic, rather than part of the theory (in this case the theory of magmas). That is why you will not see anyone listing axioms/rules for equality. Once it is specified (implicitly or not) as the first-order theory of magmas, all logical axioms/rules are automathematically included.
answered Jan 16 at 11:32
user21820user21820
39k543153
$endgroup$
2
$begingroup$
Are you sure? See here: en.wikipedia.org/wiki/Magma_(algebra)
$endgroup$
– Michael Rozenberg
Jan 16 at 13:33
1
$begingroup$
@MichaelRozenberg: I'm a logician, and it's real funny to see you citing a wikipedia article when what I'm saying is completely standard. It's clear that you don't know the difference between a theory and a model of it. Please refer to any standard introductory logic text.
$endgroup$
– user21820
Jan 16 at 13:40
add a comment |
$begingroup$
Yes, it's true by the definition of the operation.
First of all, the operation it's the function.
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Michael's answer is simply wrong. In the axiomatization of a magma $(A,+)$, the binary function-symbol $+$ is not a function but merely a symbol. Also, there are 2 main kinds of deductive-systems for first-order logic.
In a Hilbert-style system, there is only one inference rule, modus ponens, and everything else is captured by axioms. The logical axioms include axioms for equality, as Mauro pointed out.
In a natural-deduction (Gentzen-style or sequent-style or Fitch-style) system, equality is typically not captured by axioms but by inference rules.
In both kinds, equality is built into the deductive system because it is considered part of first-order logic, rather than part of the theory (in this case the theory of magmas). That is why you will not see anyone listing axioms/rules for equality. Once it is specified (implicitly or not) as the first-order theory of magmas, all logical axioms/rules are automathematically included.
answered Jan 16 at 11:32
user21820user21820
39k543153
$endgroup$
2
$begingroup$
Are you sure? See here: en.wikipedia.org/wiki/Magma_(algebra)
$endgroup$
– Michael Rozenberg
Jan 16 at 13:33
1
$begingroup$
@MichaelRozenberg: I'm a logician, and it's real funny to see you citing a wikipedia article when what I'm saying is completely standard. It's clear that you don't know the difference between a theory and a model of it. Please refer to any standard introductory logic text.
$endgroup$
– user21820
Jan 16 at 13:40
add a comment |
$begingroup$
Michael's answer is simply wrong. In the axiomatization of a magma $(A,+)$, the binary function-symbol $+$ is not a function but merely a symbol. Also, there are 2 main kinds of deductive-systems for first-order logic.
In a Hilbert-style system, there is only one inference rule, modus ponens, and everything else is captured by axioms. The logical axioms include axioms for equality, as Mauro pointed out.
In a natural-deduction (Gentzen-style or sequent-style or Fitch-style) system, equality is typically not captured by axioms but by inference rules.
In both kinds, equality is built into the deductive system because it is considered part of first-order logic, rather than part of the theory (in this case the theory of magmas). That is why you will not see anyone listing axioms/rules for equality. Once it is specified (implicitly or not) as the first-order theory of magmas, all logical axioms/rules are automathematically included.
answered Jan 16 at 11:32
user21820user21820
39k543153
$endgroup$
2
$begingroup$
Are you sure? See here: en.wikipedia.org/wiki/Magma_(algebra)
$endgroup$
– Michael Rozenberg
Jan 16 at 13:33
1
$begingroup$
@MichaelRozenberg: I'm a logician, and it's real funny to see you citing a wikipedia article when what I'm saying is completely standard. It's clear that you don't know the difference between a theory and a model of it. Please refer to any standard introductory logic text.
$endgroup$
– user21820
Jan 16 at 13:40
add a comment |
$begingroup$
Michael's answer is simply wrong. In the axiomatization of a magma $(A,+)$, the binary function-symbol $+$ is not a function but merely a symbol. Also, there are 2 main kinds of deductive-systems for first-order logic.
In a Hilbert-style system, there is only one inference rule, modus ponens, and everything else is captured by axioms. The logical axioms include axioms for equality, as Mauro pointed out.
In a natural-deduction (Gentzen-style or sequent-style or Fitch-style) system, equality is typically not captured by axioms but by inference rules.
In both kinds, equality is built into the deductive system because it is considered part of first-order logic, rather than part of the theory (in this case the theory of magmas). That is why you will not see anyone listing axioms/rules for equality. Once it is specified (implicitly or not) as the first-order theory of magmas, all logical axioms/rules are automathematically included.
answered Jan 16 at 11:32
user21820user21820
39k543153
$endgroup$
Michael's answer is simply wrong. In the axiomatization of a magma $(A,+)$, the binary function-symbol $+$ is not a function but merely a symbol. Also, there are 2 main kinds of deductive-systems for first-order logic.
In a Hilbert-style system, there is only one inference rule, modus ponens, and everything else is captured by axioms. The logical axioms include axioms for equality, as Mauro pointed out.
In a natural-deduction (Gentzen-style or sequent-style or Fitch-style) system, equality is typically not captured by axioms but by inference rules.
In both kinds, equality is built into the deductive system because it is considered part of first-order logic, rather than part of the theory (in this case the theory of magmas). That is why you will not see anyone listing axioms/rules for equality. Once it is specified (implicitly or not) as the first-order theory of magmas, all logical axioms/rules are automathematically included.
answered Jan 16 at 11:32
user21820user21820
39k543153
answered Jan 16 at 11:32
user21820user21820
39k543153
answered Jan 16 at 11:32
user21820user21820
39k543153
answered Jan 16 at 11:32
user21820user21820
39k543153
39k543153
2
$begingroup$
Are you sure? See here: en.wikipedia.org/wiki/Magma_(algebra)
$endgroup$
– Michael Rozenberg
Jan 16 at 13:33
1
$begingroup$
@MichaelRozenberg: I'm a logician, and it's real funny to see you citing a wikipedia article when what I'm saying is completely standard. It's clear that you don't know the difference between a theory and a model of it. Please refer to any standard introductory logic text.
$endgroup$
– user21820
Jan 16 at 13:40
add a comment |
2
$begingroup$
Are you sure? See here: en.wikipedia.org/wiki/Magma_(algebra)
$endgroup$
– Michael Rozenberg
Jan 16 at 13:33
1
$begingroup$
@MichaelRozenberg: I'm a logician, and it's real funny to see you citing a wikipedia article when what I'm saying is completely standard. It's clear that you don't know the difference between a theory and a model of it. Please refer to any standard introductory logic text.
$endgroup$
– user21820
Jan 16 at 13:40
2
2
$begingroup$
Are you sure? See here: en.wikipedia.org/wiki/Magma_(algebra)
$endgroup$
– Michael Rozenberg
Jan 16 at 13:33
$begingroup$
Are you sure? See here: en.wikipedia.org/wiki/Magma_(algebra)
$endgroup$
– Michael Rozenberg
Jan 16 at 13:33
1
1
$begingroup$
@MichaelRozenberg: I'm a logician, and it's real funny to see you citing a wikipedia article when what I'm saying is completely standard. It's clear that you don't know the difference between a theory and a model of it. Please refer to any standard introductory logic text.
$endgroup$
– user21820
Jan 16 at 13:40
$begingroup$
@MichaelRozenberg: I'm a logician, and it's real funny to see you citing a wikipedia article when what I'm saying is completely standard. It's clear that you don't know the difference between a theory and a model of it. Please refer to any standard introductory logic text.
$endgroup$
– user21820
Jan 16 at 13:40
add a comment |
$begingroup$
Yes, it's true by the definition of the operation.
First of all, the operation it's the function.
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
add a comment |
$begingroup$
Yes, it's true by the definition of the operation.
First of all, the operation it's the function.
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
add a comment |
$begingroup$
Yes, it's true by the definition of the operation.
First of all, the operation it's the function.
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
$endgroup$
Yes, it's true by the definition of the operation.
First of all, the operation it's the function.
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
answered Jan 15 at 8:24
Michael RozenbergMichael Rozenberg
103k1891196
103k1891196
add a comment |
add a comment |
$begingroup$
I think this is an axiom in first-order logic.
$endgroup$
– YuiTo Cheng
Jan 15 at 8:24
2
$begingroup$
It is an instance of substitution axiom for equality ; see also this post
$endgroup$
– Mauro ALLEGRANZA
Jan 15 at 8:32
2
$begingroup$
The operation is a function $+ :Atimes A to A$. If $a=b$, then $(a,c)=(b,c)$, and therefore $ + (a,c)=+ (b,c)$
$endgroup$
– leibnewtz
Jan 15 at 9:02
1
$begingroup$
Since you don't want to accept my (correct) answer, and some unknown joker downvoted it out of spite, I'm deleting it. Note that Michael Rozenberg does not understand basic first-order logic, much less the first-order theory of groups, and his answer is simply wrong but he refuses to admit it.
$endgroup$
– user21820
Feb 5 at 17:12
2
$begingroup$
@user21820 That's not a valid reason to delete an answer. Please refrain from such behaviour in the future. Your question is useful to others, and we do not encourage self-vandalism. I have upvoted your answer as a sign of good faith. Regards,
$endgroup$
– Pedro Tamaroff♦
Feb 12 at 1:04