Mixing
Intuitive notion of negation: implication example
$begingroup$
I am bit puzzled with intuitive notion of negating an implication.
Say we have an implication.
If A is true, then B is true.
Say we want to negate it. Not analytically, rather intuitively.
Now, why is following NOT a negation of implication?
If A is true, then B is either true or not (we don't know).
logic propositional-calculus
asked Jan 28 at 8:19
davidkdavidk
1
$endgroup$
add a comment |
$begingroup$
I am bit puzzled with intuitive notion of negating an implication.
Say we have an implication.
If A is true, then B is true.
Say we want to negate it. Not analytically, rather intuitively.
Now, why is following NOT a negation of implication?
If A is true, then B is either true or not (we don't know).
logic propositional-calculus
asked Jan 28 at 8:19
davidkdavidk
1
$endgroup$
$begingroup$
What you have written is a tautology. '$B$ is either true or not' is a tautology because $B$ has to be either true or false, so the statement is always true. Logically, $Boplusneg B=B+(neg B)=1$. This means you are proposing $Ato1equiv1$.
$endgroup$
– Shubham Johri
Jan 28 at 8:54
$begingroup$
@ShubhamJohri consider if A is true then B is sometimes true sometimes false, I know this kind of things might not be studied in logic, but I am trying to illustrate what I meant. That maybe negation is not only when A is true and B is false, but maybe if A is true and you don't know status of B that is also negation.
$endgroup$
– davidk
Jan 28 at 9:02
1
$begingroup$
How exactly do you define 'sometimes'? More formally, do you mean to say there is a third (or more) variable(s), say $C$, that controls the state of $B$ other than $A$?
$endgroup$
– Shubham Johri
Jan 28 at 9:13
$begingroup$
"We don't know" is open whereas in fact it is either true/false or inconsistent, or can't be known....each has its own treatement
$endgroup$
– NoChance
Jan 28 at 9:17
$begingroup$
@davidk I think you are misinterpreting negation. $neg(Ato B)$ is defined to be true whenever $Ato B$ is false and vice-versa. The trueness of the statement dictates the trueness of its negation. Stating $Ato B$ locks the trueness of the statement $neg(Ato B)$ for all the possible true-false combinations for $A,B$, leaving no room for the uncertainty 'we don't know'.
$endgroup$
– Shubham Johri
Jan 28 at 9:40
add a comment |
$begingroup$
I am bit puzzled with intuitive notion of negating an implication.
Say we have an implication.
If A is true, then B is true.
Say we want to negate it. Not analytically, rather intuitively.
Now, why is following NOT a negation of implication?
If A is true, then B is either true or not (we don't know).
logic propositional-calculus
asked Jan 28 at 8:19
davidkdavidk
1
$endgroup$
I am bit puzzled with intuitive notion of negating an implication.
Say we have an implication.
If A is true, then B is true.
Say we want to negate it. Not analytically, rather intuitively.
Now, why is following NOT a negation of implication?
If A is true, then B is either true or not (we don't know).
logic propositional-calculus
logic propositional-calculus
asked Jan 28 at 8:19
davidkdavidk
1
asked Jan 28 at 8:19
davidkdavidk
1
asked Jan 28 at 8:19
davidkdavidk
1
asked Jan 28 at 8:19
davidkdavidk
1
asked Jan 28 at 8:19
davidkdavidk
1
1
$begingroup$
What you have written is a tautology. '$B$ is either true or not' is a tautology because $B$ has to be either true or false, so the statement is always true. Logically, $Boplusneg B=B+(neg B)=1$. This means you are proposing $Ato1equiv1$.
$endgroup$
– Shubham Johri
Jan 28 at 8:54
$begingroup$
@ShubhamJohri consider if A is true then B is sometimes true sometimes false, I know this kind of things might not be studied in logic, but I am trying to illustrate what I meant. That maybe negation is not only when A is true and B is false, but maybe if A is true and you don't know status of B that is also negation.
$endgroup$
– davidk
Jan 28 at 9:02
1
$begingroup$
How exactly do you define 'sometimes'? More formally, do you mean to say there is a third (or more) variable(s), say $C$, that controls the state of $B$ other than $A$?
$endgroup$
– Shubham Johri
Jan 28 at 9:13
$begingroup$
"We don't know" is open whereas in fact it is either true/false or inconsistent, or can't be known....each has its own treatement
$endgroup$
– NoChance
Jan 28 at 9:17
$begingroup$
@davidk I think you are misinterpreting negation. $neg(Ato B)$ is defined to be true whenever $Ato B$ is false and vice-versa. The trueness of the statement dictates the trueness of its negation. Stating $Ato B$ locks the trueness of the statement $neg(Ato B)$ for all the possible true-false combinations for $A,B$, leaving no room for the uncertainty 'we don't know'.
$endgroup$
– Shubham Johri
Jan 28 at 9:40
add a comment |
$begingroup$
What you have written is a tautology. '$B$ is either true or not' is a tautology because $B$ has to be either true or false, so the statement is always true. Logically, $Boplusneg B=B+(neg B)=1$. This means you are proposing $Ato1equiv1$.
$endgroup$
– Shubham Johri
Jan 28 at 8:54
$begingroup$
@ShubhamJohri consider if A is true then B is sometimes true sometimes false, I know this kind of things might not be studied in logic, but I am trying to illustrate what I meant. That maybe negation is not only when A is true and B is false, but maybe if A is true and you don't know status of B that is also negation.
$endgroup$
– davidk
Jan 28 at 9:02
1
$begingroup$
How exactly do you define 'sometimes'? More formally, do you mean to say there is a third (or more) variable(s), say $C$, that controls the state of $B$ other than $A$?
$endgroup$
– Shubham Johri
Jan 28 at 9:13
$begingroup$
"We don't know" is open whereas in fact it is either true/false or inconsistent, or can't be known....each has its own treatement
$endgroup$
– NoChance
Jan 28 at 9:17
$begingroup$
@davidk I think you are misinterpreting negation. $neg(Ato B)$ is defined to be true whenever $Ato B$ is false and vice-versa. The trueness of the statement dictates the trueness of its negation. Stating $Ato B$ locks the trueness of the statement $neg(Ato B)$ for all the possible true-false combinations for $A,B$, leaving no room for the uncertainty 'we don't know'.
$endgroup$
– Shubham Johri
Jan 28 at 9:40
$begingroup$
What you have written is a tautology. '$B$ is either true or not' is a tautology because $B$ has to be either true or false, so the statement is always true. Logically, $Boplusneg B=B+(neg B)=1$. This means you are proposing $Ato1equiv1$.
$endgroup$
– Shubham Johri
Jan 28 at 8:54
$begingroup$
What you have written is a tautology. '$B$ is either true or not' is a tautology because $B$ has to be either true or false, so the statement is always true. Logically, $Boplusneg B=B+(neg B)=1$. This means you are proposing $Ato1equiv1$.
$endgroup$
– Shubham Johri
Jan 28 at 8:54
$begingroup$
@ShubhamJohri consider if A is true then B is sometimes true sometimes false, I know this kind of things might not be studied in logic, but I am trying to illustrate what I meant. That maybe negation is not only when A is true and B is false, but maybe if A is true and you don't know status of B that is also negation.
$endgroup$
– davidk
Jan 28 at 9:02
$begingroup$
@ShubhamJohri consider if A is true then B is sometimes true sometimes false, I know this kind of things might not be studied in logic, but I am trying to illustrate what I meant. That maybe negation is not only when A is true and B is false, but maybe if A is true and you don't know status of B that is also negation.
$endgroup$
– davidk
Jan 28 at 9:02
1
1
$begingroup$
How exactly do you define 'sometimes'? More formally, do you mean to say there is a third (or more) variable(s), say $C$, that controls the state of $B$ other than $A$?
$endgroup$
– Shubham Johri
Jan 28 at 9:13
$begingroup$
How exactly do you define 'sometimes'? More formally, do you mean to say there is a third (or more) variable(s), say $C$, that controls the state of $B$ other than $A$?
$endgroup$
– Shubham Johri
Jan 28 at 9:13
$begingroup$
"We don't know" is open whereas in fact it is either true/false or inconsistent, or can't be known....each has its own treatement
$endgroup$
– NoChance
Jan 28 at 9:17
$begingroup$
"We don't know" is open whereas in fact it is either true/false or inconsistent, or can't be known....each has its own treatement
$endgroup$
– NoChance
Jan 28 at 9:17
$begingroup$
@davidk I think you are misinterpreting negation. $neg(Ato B)$ is defined to be true whenever $Ato B$ is false and vice-versa. The trueness of the statement dictates the trueness of its negation. Stating $Ato B$ locks the trueness of the statement $neg(Ato B)$ for all the possible true-false combinations for $A,B$, leaving no room for the uncertainty 'we don't know'.
$endgroup$
– Shubham Johri
Jan 28 at 9:40
$begingroup$
@davidk I think you are misinterpreting negation. $neg(Ato B)$ is defined to be true whenever $Ato B$ is false and vice-versa. The trueness of the statement dictates the trueness of its negation. Stating $Ato B$ locks the trueness of the statement $neg(Ato B)$ for all the possible true-false combinations for $A,B$, leaving no room for the uncertainty 'we don't know'.
$endgroup$
– Shubham Johri
Jan 28 at 9:40
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The conditional $A to B$ does not mean :
"If A is true, then B is true".
The truth table for the conditional has four cases, and only one of them has FALSE as "output".
Thus, considering the negation of $A to B$, we want that it is TRUE exactly when the original one is FALSE.
I.e. $lnot (A to B)$ must be TRUE exactly when $A$ is TRUE and $B$ is FALSE.
This means that the negation of "If A is true, then B is true" is equivalent to :
"A and not B".
Another approach is : consider that $A to B$ is TRUE either when $A$ is FALSE, or when $A$ is TRUE also $B$ is.
There are many discussion about the use of conditional in natural languages and its counterpart in logic; see e.g. the so-called Paradoxes of material implication.
The Material implication of classical propositional calculus is defined through its truth table and thus it is a "simplified model" of the way natural language works.
Its usefulness in formalizing many mathematical (and not only) arguments is the only reason to use it in formal contexts.
Quite different is the case when "implication" means Logical consequence.
We have that $A vDash B$ when there is no interpretations that makes $A$ True and $B$ False.
In this case, we have that if $A$ is True, we can assert that also $B$ is.
On the other hand, if $A$ is False, then $B$ can be either True or False.
If we follow this approach, what is the negation of $A vDash B$ ?
It is :
there is at least one interpretation that makes $A$ True and $B$ False.
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
$endgroup$
$begingroup$
You answered analytically again, mine was maybe more philosophical in nature, why is: "If A is true, then B is either true or not (we don't know)." NOT a negation of implication, on a intuitive level, because it seems it could be.
$endgroup$
– davidk
Jan 28 at 8:31
$begingroup$
Very good points made here. Thank you.
$endgroup$
– NoChance
Jan 28 at 9:15
$begingroup$
The so-called paradoxes of material implications are the inevitable consequence of simple, self-evident rules of natural deduction: 1. Conditional proof ($implies$ intro), 2. Proof by contradiction ($neg$ intro), 3. Join statements ($land$ Intro), 4. Split statements ($land$ elim), 5. Double Negation ($neg neg$ elim).
$endgroup$
– Dan Christensen
Jan 28 at 13:42
add a comment |
$begingroup$
What does $A$ implies $B$ mean in classical logic? In the context of a logical proof, it means only that, at the moment, it is not the case that both A is true and B is false. Symbolically:
$A implies B space equiv space neg (A land neg B)$
This is often given as a definition in textbooks, but it can also be derived using other rules of natural deduction including other elementary properties of logical implication.
Note that there is no suggestion here of cause and effect, or the passage of time. It may help to think of it is as a single snap shot of reality. This is true of classical logic in general, and nearly all of modern mathematics. For the most part, causality and the passage of time are in the realms of science, not mathematics.
Now, the negation of $Aimplies B$ is obviously $Aland neg B$.
In words: $A$ is true and $B$ is false.
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
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add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The conditional $A to B$ does not mean :
"If A is true, then B is true".
The truth table for the conditional has four cases, and only one of them has FALSE as "output".
Thus, considering the negation of $A to B$, we want that it is TRUE exactly when the original one is FALSE.
I.e. $lnot (A to B)$ must be TRUE exactly when $A$ is TRUE and $B$ is FALSE.
This means that the negation of "If A is true, then B is true" is equivalent to :
"A and not B".
Another approach is : consider that $A to B$ is TRUE either when $A$ is FALSE, or when $A$ is TRUE also $B$ is.
There are many discussion about the use of conditional in natural languages and its counterpart in logic; see e.g. the so-called Paradoxes of material implication.
The Material implication of classical propositional calculus is defined through its truth table and thus it is a "simplified model" of the way natural language works.
Its usefulness in formalizing many mathematical (and not only) arguments is the only reason to use it in formal contexts.
Quite different is the case when "implication" means Logical consequence.
We have that $A vDash B$ when there is no interpretations that makes $A$ True and $B$ False.
In this case, we have that if $A$ is True, we can assert that also $B$ is.
On the other hand, if $A$ is False, then $B$ can be either True or False.
If we follow this approach, what is the negation of $A vDash B$ ?
It is :
there is at least one interpretation that makes $A$ True and $B$ False.
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
$endgroup$
$begingroup$
You answered analytically again, mine was maybe more philosophical in nature, why is: "If A is true, then B is either true or not (we don't know)." NOT a negation of implication, on a intuitive level, because it seems it could be.
$endgroup$
– davidk
Jan 28 at 8:31
$begingroup$
Very good points made here. Thank you.
$endgroup$
– NoChance
Jan 28 at 9:15
$begingroup$
The so-called paradoxes of material implications are the inevitable consequence of simple, self-evident rules of natural deduction: 1. Conditional proof ($implies$ intro), 2. Proof by contradiction ($neg$ intro), 3. Join statements ($land$ Intro), 4. Split statements ($land$ elim), 5. Double Negation ($neg neg$ elim).
$endgroup$
– Dan Christensen
Jan 28 at 13:42
add a comment |
$begingroup$
The conditional $A to B$ does not mean :
"If A is true, then B is true".
The truth table for the conditional has four cases, and only one of them has FALSE as "output".
Thus, considering the negation of $A to B$, we want that it is TRUE exactly when the original one is FALSE.
I.e. $lnot (A to B)$ must be TRUE exactly when $A$ is TRUE and $B$ is FALSE.
This means that the negation of "If A is true, then B is true" is equivalent to :
"A and not B".
Another approach is : consider that $A to B$ is TRUE either when $A$ is FALSE, or when $A$ is TRUE also $B$ is.
There are many discussion about the use of conditional in natural languages and its counterpart in logic; see e.g. the so-called Paradoxes of material implication.
The Material implication of classical propositional calculus is defined through its truth table and thus it is a "simplified model" of the way natural language works.
Its usefulness in formalizing many mathematical (and not only) arguments is the only reason to use it in formal contexts.
Quite different is the case when "implication" means Logical consequence.
We have that $A vDash B$ when there is no interpretations that makes $A$ True and $B$ False.
In this case, we have that if $A$ is True, we can assert that also $B$ is.
On the other hand, if $A$ is False, then $B$ can be either True or False.
If we follow this approach, what is the negation of $A vDash B$ ?
It is :
there is at least one interpretation that makes $A$ True and $B$ False.
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
$endgroup$
$begingroup$
You answered analytically again, mine was maybe more philosophical in nature, why is: "If A is true, then B is either true or not (we don't know)." NOT a negation of implication, on a intuitive level, because it seems it could be.
$endgroup$
– davidk
Jan 28 at 8:31
$begingroup$
Very good points made here. Thank you.
$endgroup$
– NoChance
Jan 28 at 9:15
$begingroup$
The so-called paradoxes of material implications are the inevitable consequence of simple, self-evident rules of natural deduction: 1. Conditional proof ($implies$ intro), 2. Proof by contradiction ($neg$ intro), 3. Join statements ($land$ Intro), 4. Split statements ($land$ elim), 5. Double Negation ($neg neg$ elim).
$endgroup$
– Dan Christensen
Jan 28 at 13:42
add a comment |
$begingroup$
The conditional $A to B$ does not mean :
"If A is true, then B is true".
The truth table for the conditional has four cases, and only one of them has FALSE as "output".
Thus, considering the negation of $A to B$, we want that it is TRUE exactly when the original one is FALSE.
I.e. $lnot (A to B)$ must be TRUE exactly when $A$ is TRUE and $B$ is FALSE.
This means that the negation of "If A is true, then B is true" is equivalent to :
"A and not B".
Another approach is : consider that $A to B$ is TRUE either when $A$ is FALSE, or when $A$ is TRUE also $B$ is.
There are many discussion about the use of conditional in natural languages and its counterpart in logic; see e.g. the so-called Paradoxes of material implication.
The Material implication of classical propositional calculus is defined through its truth table and thus it is a "simplified model" of the way natural language works.
Its usefulness in formalizing many mathematical (and not only) arguments is the only reason to use it in formal contexts.
Quite different is the case when "implication" means Logical consequence.
We have that $A vDash B$ when there is no interpretations that makes $A$ True and $B$ False.
In this case, we have that if $A$ is True, we can assert that also $B$ is.
On the other hand, if $A$ is False, then $B$ can be either True or False.
If we follow this approach, what is the negation of $A vDash B$ ?
It is :
there is at least one interpretation that makes $A$ True and $B$ False.
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
$endgroup$
The conditional $A to B$ does not mean :
"If A is true, then B is true".
The truth table for the conditional has four cases, and only one of them has FALSE as "output".
Thus, considering the negation of $A to B$, we want that it is TRUE exactly when the original one is FALSE.
I.e. $lnot (A to B)$ must be TRUE exactly when $A$ is TRUE and $B$ is FALSE.
This means that the negation of "If A is true, then B is true" is equivalent to :
"A and not B".
Another approach is : consider that $A to B$ is TRUE either when $A$ is FALSE, or when $A$ is TRUE also $B$ is.
There are many discussion about the use of conditional in natural languages and its counterpart in logic; see e.g. the so-called Paradoxes of material implication.
The Material implication of classical propositional calculus is defined through its truth table and thus it is a "simplified model" of the way natural language works.
Its usefulness in formalizing many mathematical (and not only) arguments is the only reason to use it in formal contexts.
Quite different is the case when "implication" means Logical consequence.
We have that $A vDash B$ when there is no interpretations that makes $A$ True and $B$ False.
In this case, we have that if $A$ is True, we can assert that also $B$ is.
On the other hand, if $A$ is False, then $B$ can be either True or False.
If we follow this approach, what is the negation of $A vDash B$ ?
It is :
there is at least one interpretation that makes $A$ True and $B$ False.
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
edited Jan 28 at 11:52
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
answered Jan 28 at 8:26
Mauro ALLEGRANZAMauro ALLEGRANZA
67.5k449117
67.5k449117
$begingroup$
You answered analytically again, mine was maybe more philosophical in nature, why is: "If A is true, then B is either true or not (we don't know)." NOT a negation of implication, on a intuitive level, because it seems it could be.
$endgroup$
– davidk
Jan 28 at 8:31
$begingroup$
Very good points made here. Thank you.
$endgroup$
– NoChance
Jan 28 at 9:15
$begingroup$
The so-called paradoxes of material implications are the inevitable consequence of simple, self-evident rules of natural deduction: 1. Conditional proof ($implies$ intro), 2. Proof by contradiction ($neg$ intro), 3. Join statements ($land$ Intro), 4. Split statements ($land$ elim), 5. Double Negation ($neg neg$ elim).
$endgroup$
– Dan Christensen
Jan 28 at 13:42
add a comment |
$begingroup$
You answered analytically again, mine was maybe more philosophical in nature, why is: "If A is true, then B is either true or not (we don't know)." NOT a negation of implication, on a intuitive level, because it seems it could be.
$endgroup$
– davidk
Jan 28 at 8:31
$begingroup$
Very good points made here. Thank you.
$endgroup$
– NoChance
Jan 28 at 9:15
$begingroup$
The so-called paradoxes of material implications are the inevitable consequence of simple, self-evident rules of natural deduction: 1. Conditional proof ($implies$ intro), 2. Proof by contradiction ($neg$ intro), 3. Join statements ($land$ Intro), 4. Split statements ($land$ elim), 5. Double Negation ($neg neg$ elim).
$endgroup$
– Dan Christensen
Jan 28 at 13:42
$begingroup$
You answered analytically again, mine was maybe more philosophical in nature, why is: "If A is true, then B is either true or not (we don't know)." NOT a negation of implication, on a intuitive level, because it seems it could be.
$endgroup$
– davidk
Jan 28 at 8:31
$begingroup$
You answered analytically again, mine was maybe more philosophical in nature, why is: "If A is true, then B is either true or not (we don't know)." NOT a negation of implication, on a intuitive level, because it seems it could be.
$endgroup$
– davidk
Jan 28 at 8:31
$begingroup$
Very good points made here. Thank you.
$endgroup$
– NoChance
Jan 28 at 9:15
$begingroup$
Very good points made here. Thank you.
$endgroup$
– NoChance
Jan 28 at 9:15
$begingroup$
The so-called paradoxes of material implications are the inevitable consequence of simple, self-evident rules of natural deduction: 1. Conditional proof ($implies$ intro), 2. Proof by contradiction ($neg$ intro), 3. Join statements ($land$ Intro), 4. Split statements ($land$ elim), 5. Double Negation ($neg neg$ elim).
$endgroup$
– Dan Christensen
Jan 28 at 13:42
$begingroup$
The so-called paradoxes of material implications are the inevitable consequence of simple, self-evident rules of natural deduction: 1. Conditional proof ($implies$ intro), 2. Proof by contradiction ($neg$ intro), 3. Join statements ($land$ Intro), 4. Split statements ($land$ elim), 5. Double Negation ($neg neg$ elim).
$endgroup$
– Dan Christensen
Jan 28 at 13:42
add a comment |
$begingroup$
What does $A$ implies $B$ mean in classical logic? In the context of a logical proof, it means only that, at the moment, it is not the case that both A is true and B is false. Symbolically:
$A implies B space equiv space neg (A land neg B)$
This is often given as a definition in textbooks, but it can also be derived using other rules of natural deduction including other elementary properties of logical implication.
Note that there is no suggestion here of cause and effect, or the passage of time. It may help to think of it is as a single snap shot of reality. This is true of classical logic in general, and nearly all of modern mathematics. For the most part, causality and the passage of time are in the realms of science, not mathematics.
Now, the negation of $Aimplies B$ is obviously $Aland neg B$.
In words: $A$ is true and $B$ is false.
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
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add a comment |
$begingroup$
What does $A$ implies $B$ mean in classical logic? In the context of a logical proof, it means only that, at the moment, it is not the case that both A is true and B is false. Symbolically:
$A implies B space equiv space neg (A land neg B)$
This is often given as a definition in textbooks, but it can also be derived using other rules of natural deduction including other elementary properties of logical implication.
Note that there is no suggestion here of cause and effect, or the passage of time. It may help to think of it is as a single snap shot of reality. This is true of classical logic in general, and nearly all of modern mathematics. For the most part, causality and the passage of time are in the realms of science, not mathematics.
Now, the negation of $Aimplies B$ is obviously $Aland neg B$.
In words: $A$ is true and $B$ is false.
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
$endgroup$
add a comment |
$begingroup$
What does $A$ implies $B$ mean in classical logic? In the context of a logical proof, it means only that, at the moment, it is not the case that both A is true and B is false. Symbolically:
$A implies B space equiv space neg (A land neg B)$
This is often given as a definition in textbooks, but it can also be derived using other rules of natural deduction including other elementary properties of logical implication.
Note that there is no suggestion here of cause and effect, or the passage of time. It may help to think of it is as a single snap shot of reality. This is true of classical logic in general, and nearly all of modern mathematics. For the most part, causality and the passage of time are in the realms of science, not mathematics.
Now, the negation of $Aimplies B$ is obviously $Aland neg B$.
In words: $A$ is true and $B$ is false.
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
$endgroup$
What does $A$ implies $B$ mean in classical logic? In the context of a logical proof, it means only that, at the moment, it is not the case that both A is true and B is false. Symbolically:
$A implies B space equiv space neg (A land neg B)$
This is often given as a definition in textbooks, but it can also be derived using other rules of natural deduction including other elementary properties of logical implication.
Note that there is no suggestion here of cause and effect, or the passage of time. It may help to think of it is as a single snap shot of reality. This is true of classical logic in general, and nearly all of modern mathematics. For the most part, causality and the passage of time are in the realms of science, not mathematics.
Now, the negation of $Aimplies B$ is obviously $Aland neg B$.
In words: $A$ is true and $B$ is false.
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
edited Jan 28 at 18:15
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
answered Jan 28 at 13:27
Dan ChristensenDan Christensen
8,66321835
8,66321835
add a comment |
add a comment |
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What you have written is a tautology. '$B$ is either true or not' is a tautology because $B$ has to be either true or false, so the statement is always true. Logically, $Boplusneg B=B+(neg B)=1$. This means you are proposing $Ato1equiv1$.
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– Shubham Johri
Jan 28 at 8:54
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@ShubhamJohri consider if A is true then B is sometimes true sometimes false, I know this kind of things might not be studied in logic, but I am trying to illustrate what I meant. That maybe negation is not only when A is true and B is false, but maybe if A is true and you don't know status of B that is also negation.
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– davidk
Jan 28 at 9:02
1
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How exactly do you define 'sometimes'? More formally, do you mean to say there is a third (or more) variable(s), say $C$, that controls the state of $B$ other than $A$?
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– Shubham Johri
Jan 28 at 9:13
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"We don't know" is open whereas in fact it is either true/false or inconsistent, or can't be known....each has its own treatement
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– NoChance
Jan 28 at 9:17
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@davidk I think you are misinterpreting negation. $neg(Ato B)$ is defined to be true whenever $Ato B$ is false and vice-versa. The trueness of the statement dictates the trueness of its negation. Stating $Ato B$ locks the trueness of the statement $neg(Ato B)$ for all the possible true-false combinations for $A,B$, leaving no room for the uncertainty 'we don't know'.
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– Shubham Johri
Jan 28 at 9:40