Why isn’t ‘because’ a logical connective in propositional logic?
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In simple terms, could someone explain why there is not a logical connective for ‘because’ in propositional logic like there is for ‘and’ and ‘or’?
Is this because the equivalent of ‘because’ is the argument of the form ‘if p, then q’, or am I missing something?
Please illustrate your answer with example(s) if possible.
logic propositional-calculus
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add a comment |
$begingroup$
In simple terms, could someone explain why there is not a logical connective for ‘because’ in propositional logic like there is for ‘and’ and ‘or’?
Is this because the equivalent of ‘because’ is the argument of the form ‘if p, then q’, or am I missing something?
Please illustrate your answer with example(s) if possible.
logic propositional-calculus
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16
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"because" is about causality, not implication
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– Hagen von Eitzen
Nov 11 '18 at 13:17
1
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Folks might find basicbooks.com/titles/judea-pearl/the-book-of-why/9780465097609 of interest.
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– Barry Cipra
Nov 11 '18 at 13:23
1
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I would love to see a complete prepositional logic theory where P→Q is not equivalent to ¬P∨Q but I haven't yet. :(
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– Joshua
Nov 12 '18 at 19:10
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@Joshua Maybe I'm misunderstanding what you mean, but that equivalence typically fails in non-classical logics. It definitely fails in constructive/intuitionistic logic. So Intuitionistic Propositional Logic would be a propositional logic where $Pto Q$ is not equivalent to $neg Plor Q$.
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– Derek Elkins
Nov 13 '18 at 6:23
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$(Pto Q) lor (Qto P)$ is a tautology in classical logic, so presumably $(Qleftarrow P) lor (Pleftarrow Q)$ is too. You would not want to read "$leftarrow$" as because here
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– Henry
Nov 13 '18 at 8:41
add a comment |
$begingroup$
In simple terms, could someone explain why there is not a logical connective for ‘because’ in propositional logic like there is for ‘and’ and ‘or’?
Is this because the equivalent of ‘because’ is the argument of the form ‘if p, then q’, or am I missing something?
Please illustrate your answer with example(s) if possible.
logic propositional-calculus
$endgroup$
In simple terms, could someone explain why there is not a logical connective for ‘because’ in propositional logic like there is for ‘and’ and ‘or’?
Is this because the equivalent of ‘because’ is the argument of the form ‘if p, then q’, or am I missing something?
Please illustrate your answer with example(s) if possible.
logic propositional-calculus
logic propositional-calculus
asked Nov 11 '18 at 13:10
seekerseeker
2,72965285
2,72965285
16
$begingroup$
"because" is about causality, not implication
$endgroup$
– Hagen von Eitzen
Nov 11 '18 at 13:17
1
$begingroup$
Folks might find basicbooks.com/titles/judea-pearl/the-book-of-why/9780465097609 of interest.
$endgroup$
– Barry Cipra
Nov 11 '18 at 13:23
1
$begingroup$
I would love to see a complete prepositional logic theory where P→Q is not equivalent to ¬P∨Q but I haven't yet. :(
$endgroup$
– Joshua
Nov 12 '18 at 19:10
$begingroup$
@Joshua Maybe I'm misunderstanding what you mean, but that equivalence typically fails in non-classical logics. It definitely fails in constructive/intuitionistic logic. So Intuitionistic Propositional Logic would be a propositional logic where $Pto Q$ is not equivalent to $neg Plor Q$.
$endgroup$
– Derek Elkins
Nov 13 '18 at 6:23
$begingroup$
$(Pto Q) lor (Qto P)$ is a tautology in classical logic, so presumably $(Qleftarrow P) lor (Pleftarrow Q)$ is too. You would not want to read "$leftarrow$" as because here
$endgroup$
– Henry
Nov 13 '18 at 8:41
add a comment |
16
$begingroup$
"because" is about causality, not implication
$endgroup$
– Hagen von Eitzen
Nov 11 '18 at 13:17
1
$begingroup$
Folks might find basicbooks.com/titles/judea-pearl/the-book-of-why/9780465097609 of interest.
$endgroup$
– Barry Cipra
Nov 11 '18 at 13:23
1
$begingroup$
I would love to see a complete prepositional logic theory where P→Q is not equivalent to ¬P∨Q but I haven't yet. :(
$endgroup$
– Joshua
Nov 12 '18 at 19:10
$begingroup$
@Joshua Maybe I'm misunderstanding what you mean, but that equivalence typically fails in non-classical logics. It definitely fails in constructive/intuitionistic logic. So Intuitionistic Propositional Logic would be a propositional logic where $Pto Q$ is not equivalent to $neg Plor Q$.
$endgroup$
– Derek Elkins
Nov 13 '18 at 6:23
$begingroup$
$(Pto Q) lor (Qto P)$ is a tautology in classical logic, so presumably $(Qleftarrow P) lor (Pleftarrow Q)$ is too. You would not want to read "$leftarrow$" as because here
$endgroup$
– Henry
Nov 13 '18 at 8:41
16
16
$begingroup$
"because" is about causality, not implication
$endgroup$
– Hagen von Eitzen
Nov 11 '18 at 13:17
$begingroup$
"because" is about causality, not implication
$endgroup$
– Hagen von Eitzen
Nov 11 '18 at 13:17
1
1
$begingroup$
Folks might find basicbooks.com/titles/judea-pearl/the-book-of-why/9780465097609 of interest.
$endgroup$
– Barry Cipra
Nov 11 '18 at 13:23
$begingroup$
Folks might find basicbooks.com/titles/judea-pearl/the-book-of-why/9780465097609 of interest.
$endgroup$
– Barry Cipra
Nov 11 '18 at 13:23
1
1
$begingroup$
I would love to see a complete prepositional logic theory where P→Q is not equivalent to ¬P∨Q but I haven't yet. :(
$endgroup$
– Joshua
Nov 12 '18 at 19:10
$begingroup$
I would love to see a complete prepositional logic theory where P→Q is not equivalent to ¬P∨Q but I haven't yet. :(
$endgroup$
– Joshua
Nov 12 '18 at 19:10
$begingroup$
@Joshua Maybe I'm misunderstanding what you mean, but that equivalence typically fails in non-classical logics. It definitely fails in constructive/intuitionistic logic. So Intuitionistic Propositional Logic would be a propositional logic where $Pto Q$ is not equivalent to $neg Plor Q$.
$endgroup$
– Derek Elkins
Nov 13 '18 at 6:23
$begingroup$
@Joshua Maybe I'm misunderstanding what you mean, but that equivalence typically fails in non-classical logics. It definitely fails in constructive/intuitionistic logic. So Intuitionistic Propositional Logic would be a propositional logic where $Pto Q$ is not equivalent to $neg Plor Q$.
$endgroup$
– Derek Elkins
Nov 13 '18 at 6:23
$begingroup$
$(Pto Q) lor (Qto P)$ is a tautology in classical logic, so presumably $(Qleftarrow P) lor (Pleftarrow Q)$ is too. You would not want to read "$leftarrow$" as because here
$endgroup$
– Henry
Nov 13 '18 at 8:41
$begingroup$
$(Pto Q) lor (Qto P)$ is a tautology in classical logic, so presumably $(Qleftarrow P) lor (Pleftarrow Q)$ is too. You would not want to read "$leftarrow$" as because here
$endgroup$
– Henry
Nov 13 '18 at 8:41
add a comment |
4 Answers
4
active
oldest
votes
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It is because 'because' is not truth-functional.
That is, knowing the truth-values of $P$ and $Q$ does not tell you the truth-value of '$P$ because of $Q$'
For example, the two statements 'Grass is green' and 'Snow is white' are both true, but 'Grass is green because snow is white' is an invalid argument, and hence, as a statement as to the validity of that argument, a false statement.
On the other hand,'Grass is green because grass is green' is a true statement as to the validity of this as an argument, but yet again it involves two true statements.
This shows that with $P$ and $Q$ both being true, the statement '$P$ because of $Q$' can either be true or false, and hence it is not truth-functional.
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12
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This is an exceedingly better answer than I expected possible. Kudos!
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– Daniel R. Collins
Nov 11 '18 at 20:25
3
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Wouldn't this reaoning also apply to if statements? "If snow is white, then grass is green" happens to be true, despite not making much sense in common English.
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– Vaelus
Nov 12 '18 at 7:38
4
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@Vaelus If I've understood the answer correctly, the point is that the truth of "if P then Q" is uniquely determined by the truth values of P and Q, whereas the truth of "P because Q" is not determined by the truth values of P and Q. That is, there are cases where "[true statement] because [true statement]" is true, and also cases where "[true statement] because [true statement]" is false, which is not the case for logical if statements.
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– David Z
Nov 12 '18 at 9:52
5
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@Vaelus Yes, the English 'if ... then ...' does not seem to be teuth-functional either .... so why do logicians define a truth-functional operator (called the material implication) to try and capture it? There is a long standing debate about this. Please look up 'Paradox of Material Implication' if you want to know more.
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– Bram28
Nov 12 '18 at 12:19
1
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@Ooker No, "Q happens because P happens" is not what he is saying, he is saying "(I know Q is true) because (I know P is true)". This is not stating anything about causality. For example, in a simple world it may be true that "If it is wet outside today, it rained last night." (and also "If it rained last night, it is wet outside today"). But only one of "It is wet outside today because it rained last night" and "It rained last night because it is wet outside today" is true, namely rain => wet. It is however true that I know it rained because I know it is wet.
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– kutschkem
Nov 13 '18 at 14:40
|
show 7 more comments
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Why isn’t ‘because’ a logical connective in propositional logic?
Is this because the equivalent of ‘because’ is the argument of the form ‘if $p$, then $q$’ ?
Exactly.
Either the connective "because" is truth-functional, in which case it is the same as "if..., then...", or it is not truth-functional, in which case we need a different way of modelling it.
See e.g. Counterfactual Theories of Causation.
See also Arthur Burks, The Logic of Causal Proposition, Mind (1951).
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I would have thought a truth-functional because might be considered the reverse of if..., then... rather than the same
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– Henry
Nov 13 '18 at 8:46
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@Henry - reverse ?
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– Mauro ALLEGRANZA
Nov 13 '18 at 9:16
2
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What does the first word (exactly) refer to?
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– Carsten S
Nov 13 '18 at 10:40
add a comment |
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I agree with the other answers, however I want to add that the closest thing might be the turnstyle symbol $vdash$, although this is usually read as "yields", and thus points the other way. If I write
$$A vdash B$$
this is read as "A yields B", or "knowing A, I can prove B". If you wanted to encode because, you could probably read it backwards as "B because of A".
Note however that this is not used as part of a logical formula, but as a shorthand between formulas when writing down a proof. So $A vdash B$ is no longer a formula, but rather a statement on how to prove $B$. (In most of the rest of mathematics, you would write $Rightarrow$ in your proof instead, however in logic this is of course easily confused with the implication inside formulas)
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2
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Yes, this would therefore be a 'meta-logical' symbol;: a symbol used to say something about logic exprssion .. but it is not a logical connective or operator.
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– Bram28
Nov 12 '18 at 12:21
add a comment |
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You can define things however you want. (Be careful; you may accidentally be inconsistent.)
Either:
- "because" is logically equivalent to a binary operator
- or it's not
If it is, it's probably the same as "only if" (Or take your pick of the other 15 operators). Adding a "because" overload would create an additional word to remember: unneeded complexity. We like simplicity.
If it isn't, you can define a binary function because(a, b)
however you want.
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add a comment |
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4 Answers
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active
oldest
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4 Answers
4
active
oldest
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$begingroup$
It is because 'because' is not truth-functional.
That is, knowing the truth-values of $P$ and $Q$ does not tell you the truth-value of '$P$ because of $Q$'
For example, the two statements 'Grass is green' and 'Snow is white' are both true, but 'Grass is green because snow is white' is an invalid argument, and hence, as a statement as to the validity of that argument, a false statement.
On the other hand,'Grass is green because grass is green' is a true statement as to the validity of this as an argument, but yet again it involves two true statements.
This shows that with $P$ and $Q$ both being true, the statement '$P$ because of $Q$' can either be true or false, and hence it is not truth-functional.
$endgroup$
12
$begingroup$
This is an exceedingly better answer than I expected possible. Kudos!
$endgroup$
– Daniel R. Collins
Nov 11 '18 at 20:25
3
$begingroup$
Wouldn't this reaoning also apply to if statements? "If snow is white, then grass is green" happens to be true, despite not making much sense in common English.
$endgroup$
– Vaelus
Nov 12 '18 at 7:38
4
$begingroup$
@Vaelus If I've understood the answer correctly, the point is that the truth of "if P then Q" is uniquely determined by the truth values of P and Q, whereas the truth of "P because Q" is not determined by the truth values of P and Q. That is, there are cases where "[true statement] because [true statement]" is true, and also cases where "[true statement] because [true statement]" is false, which is not the case for logical if statements.
$endgroup$
– David Z
Nov 12 '18 at 9:52
5
$begingroup$
@Vaelus Yes, the English 'if ... then ...' does not seem to be teuth-functional either .... so why do logicians define a truth-functional operator (called the material implication) to try and capture it? There is a long standing debate about this. Please look up 'Paradox of Material Implication' if you want to know more.
$endgroup$
– Bram28
Nov 12 '18 at 12:19
1
$begingroup$
@Ooker No, "Q happens because P happens" is not what he is saying, he is saying "(I know Q is true) because (I know P is true)". This is not stating anything about causality. For example, in a simple world it may be true that "If it is wet outside today, it rained last night." (and also "If it rained last night, it is wet outside today"). But only one of "It is wet outside today because it rained last night" and "It rained last night because it is wet outside today" is true, namely rain => wet. It is however true that I know it rained because I know it is wet.
$endgroup$
– kutschkem
Nov 13 '18 at 14:40
|
show 7 more comments
$begingroup$
It is because 'because' is not truth-functional.
That is, knowing the truth-values of $P$ and $Q$ does not tell you the truth-value of '$P$ because of $Q$'
For example, the two statements 'Grass is green' and 'Snow is white' are both true, but 'Grass is green because snow is white' is an invalid argument, and hence, as a statement as to the validity of that argument, a false statement.
On the other hand,'Grass is green because grass is green' is a true statement as to the validity of this as an argument, but yet again it involves two true statements.
This shows that with $P$ and $Q$ both being true, the statement '$P$ because of $Q$' can either be true or false, and hence it is not truth-functional.
$endgroup$
12
$begingroup$
This is an exceedingly better answer than I expected possible. Kudos!
$endgroup$
– Daniel R. Collins
Nov 11 '18 at 20:25
3
$begingroup$
Wouldn't this reaoning also apply to if statements? "If snow is white, then grass is green" happens to be true, despite not making much sense in common English.
$endgroup$
– Vaelus
Nov 12 '18 at 7:38
4
$begingroup$
@Vaelus If I've understood the answer correctly, the point is that the truth of "if P then Q" is uniquely determined by the truth values of P and Q, whereas the truth of "P because Q" is not determined by the truth values of P and Q. That is, there are cases where "[true statement] because [true statement]" is true, and also cases where "[true statement] because [true statement]" is false, which is not the case for logical if statements.
$endgroup$
– David Z
Nov 12 '18 at 9:52
5
$begingroup$
@Vaelus Yes, the English 'if ... then ...' does not seem to be teuth-functional either .... so why do logicians define a truth-functional operator (called the material implication) to try and capture it? There is a long standing debate about this. Please look up 'Paradox of Material Implication' if you want to know more.
$endgroup$
– Bram28
Nov 12 '18 at 12:19
1
$begingroup$
@Ooker No, "Q happens because P happens" is not what he is saying, he is saying "(I know Q is true) because (I know P is true)". This is not stating anything about causality. For example, in a simple world it may be true that "If it is wet outside today, it rained last night." (and also "If it rained last night, it is wet outside today"). But only one of "It is wet outside today because it rained last night" and "It rained last night because it is wet outside today" is true, namely rain => wet. It is however true that I know it rained because I know it is wet.
$endgroup$
– kutschkem
Nov 13 '18 at 14:40
|
show 7 more comments
$begingroup$
It is because 'because' is not truth-functional.
That is, knowing the truth-values of $P$ and $Q$ does not tell you the truth-value of '$P$ because of $Q$'
For example, the two statements 'Grass is green' and 'Snow is white' are both true, but 'Grass is green because snow is white' is an invalid argument, and hence, as a statement as to the validity of that argument, a false statement.
On the other hand,'Grass is green because grass is green' is a true statement as to the validity of this as an argument, but yet again it involves two true statements.
This shows that with $P$ and $Q$ both being true, the statement '$P$ because of $Q$' can either be true or false, and hence it is not truth-functional.
$endgroup$
It is because 'because' is not truth-functional.
That is, knowing the truth-values of $P$ and $Q$ does not tell you the truth-value of '$P$ because of $Q$'
For example, the two statements 'Grass is green' and 'Snow is white' are both true, but 'Grass is green because snow is white' is an invalid argument, and hence, as a statement as to the validity of that argument, a false statement.
On the other hand,'Grass is green because grass is green' is a true statement as to the validity of this as an argument, but yet again it involves two true statements.
This shows that with $P$ and $Q$ both being true, the statement '$P$ because of $Q$' can either be true or false, and hence it is not truth-functional.
answered Nov 11 '18 at 13:36
Bram28Bram28
61.9k44793
61.9k44793
12
$begingroup$
This is an exceedingly better answer than I expected possible. Kudos!
$endgroup$
– Daniel R. Collins
Nov 11 '18 at 20:25
3
$begingroup$
Wouldn't this reaoning also apply to if statements? "If snow is white, then grass is green" happens to be true, despite not making much sense in common English.
$endgroup$
– Vaelus
Nov 12 '18 at 7:38
4
$begingroup$
@Vaelus If I've understood the answer correctly, the point is that the truth of "if P then Q" is uniquely determined by the truth values of P and Q, whereas the truth of "P because Q" is not determined by the truth values of P and Q. That is, there are cases where "[true statement] because [true statement]" is true, and also cases where "[true statement] because [true statement]" is false, which is not the case for logical if statements.
$endgroup$
– David Z
Nov 12 '18 at 9:52
5
$begingroup$
@Vaelus Yes, the English 'if ... then ...' does not seem to be teuth-functional either .... so why do logicians define a truth-functional operator (called the material implication) to try and capture it? There is a long standing debate about this. Please look up 'Paradox of Material Implication' if you want to know more.
$endgroup$
– Bram28
Nov 12 '18 at 12:19
1
$begingroup$
@Ooker No, "Q happens because P happens" is not what he is saying, he is saying "(I know Q is true) because (I know P is true)". This is not stating anything about causality. For example, in a simple world it may be true that "If it is wet outside today, it rained last night." (and also "If it rained last night, it is wet outside today"). But only one of "It is wet outside today because it rained last night" and "It rained last night because it is wet outside today" is true, namely rain => wet. It is however true that I know it rained because I know it is wet.
$endgroup$
– kutschkem
Nov 13 '18 at 14:40
|
show 7 more comments
12
$begingroup$
This is an exceedingly better answer than I expected possible. Kudos!
$endgroup$
– Daniel R. Collins
Nov 11 '18 at 20:25
3
$begingroup$
Wouldn't this reaoning also apply to if statements? "If snow is white, then grass is green" happens to be true, despite not making much sense in common English.
$endgroup$
– Vaelus
Nov 12 '18 at 7:38
4
$begingroup$
@Vaelus If I've understood the answer correctly, the point is that the truth of "if P then Q" is uniquely determined by the truth values of P and Q, whereas the truth of "P because Q" is not determined by the truth values of P and Q. That is, there are cases where "[true statement] because [true statement]" is true, and also cases where "[true statement] because [true statement]" is false, which is not the case for logical if statements.
$endgroup$
– David Z
Nov 12 '18 at 9:52
5
$begingroup$
@Vaelus Yes, the English 'if ... then ...' does not seem to be teuth-functional either .... so why do logicians define a truth-functional operator (called the material implication) to try and capture it? There is a long standing debate about this. Please look up 'Paradox of Material Implication' if you want to know more.
$endgroup$
– Bram28
Nov 12 '18 at 12:19
1
$begingroup$
@Ooker No, "Q happens because P happens" is not what he is saying, he is saying "(I know Q is true) because (I know P is true)". This is not stating anything about causality. For example, in a simple world it may be true that "If it is wet outside today, it rained last night." (and also "If it rained last night, it is wet outside today"). But only one of "It is wet outside today because it rained last night" and "It rained last night because it is wet outside today" is true, namely rain => wet. It is however true that I know it rained because I know it is wet.
$endgroup$
– kutschkem
Nov 13 '18 at 14:40
12
12
$begingroup$
This is an exceedingly better answer than I expected possible. Kudos!
$endgroup$
– Daniel R. Collins
Nov 11 '18 at 20:25
$begingroup$
This is an exceedingly better answer than I expected possible. Kudos!
$endgroup$
– Daniel R. Collins
Nov 11 '18 at 20:25
3
3
$begingroup$
Wouldn't this reaoning also apply to if statements? "If snow is white, then grass is green" happens to be true, despite not making much sense in common English.
$endgroup$
– Vaelus
Nov 12 '18 at 7:38
$begingroup$
Wouldn't this reaoning also apply to if statements? "If snow is white, then grass is green" happens to be true, despite not making much sense in common English.
$endgroup$
– Vaelus
Nov 12 '18 at 7:38
4
4
$begingroup$
@Vaelus If I've understood the answer correctly, the point is that the truth of "if P then Q" is uniquely determined by the truth values of P and Q, whereas the truth of "P because Q" is not determined by the truth values of P and Q. That is, there are cases where "[true statement] because [true statement]" is true, and also cases where "[true statement] because [true statement]" is false, which is not the case for logical if statements.
$endgroup$
– David Z
Nov 12 '18 at 9:52
$begingroup$
@Vaelus If I've understood the answer correctly, the point is that the truth of "if P then Q" is uniquely determined by the truth values of P and Q, whereas the truth of "P because Q" is not determined by the truth values of P and Q. That is, there are cases where "[true statement] because [true statement]" is true, and also cases where "[true statement] because [true statement]" is false, which is not the case for logical if statements.
$endgroup$
– David Z
Nov 12 '18 at 9:52
5
5
$begingroup$
@Vaelus Yes, the English 'if ... then ...' does not seem to be teuth-functional either .... so why do logicians define a truth-functional operator (called the material implication) to try and capture it? There is a long standing debate about this. Please look up 'Paradox of Material Implication' if you want to know more.
$endgroup$
– Bram28
Nov 12 '18 at 12:19
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@Vaelus Yes, the English 'if ... then ...' does not seem to be teuth-functional either .... so why do logicians define a truth-functional operator (called the material implication) to try and capture it? There is a long standing debate about this. Please look up 'Paradox of Material Implication' if you want to know more.
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– Bram28
Nov 12 '18 at 12:19
1
1
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@Ooker No, "Q happens because P happens" is not what he is saying, he is saying "(I know Q is true) because (I know P is true)". This is not stating anything about causality. For example, in a simple world it may be true that "If it is wet outside today, it rained last night." (and also "If it rained last night, it is wet outside today"). But only one of "It is wet outside today because it rained last night" and "It rained last night because it is wet outside today" is true, namely rain => wet. It is however true that I know it rained because I know it is wet.
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– kutschkem
Nov 13 '18 at 14:40
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@Ooker No, "Q happens because P happens" is not what he is saying, he is saying "(I know Q is true) because (I know P is true)". This is not stating anything about causality. For example, in a simple world it may be true that "If it is wet outside today, it rained last night." (and also "If it rained last night, it is wet outside today"). But only one of "It is wet outside today because it rained last night" and "It rained last night because it is wet outside today" is true, namely rain => wet. It is however true that I know it rained because I know it is wet.
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– kutschkem
Nov 13 '18 at 14:40
|
show 7 more comments
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Why isn’t ‘because’ a logical connective in propositional logic?
Is this because the equivalent of ‘because’ is the argument of the form ‘if $p$, then $q$’ ?
Exactly.
Either the connective "because" is truth-functional, in which case it is the same as "if..., then...", or it is not truth-functional, in which case we need a different way of modelling it.
See e.g. Counterfactual Theories of Causation.
See also Arthur Burks, The Logic of Causal Proposition, Mind (1951).
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I would have thought a truth-functional because might be considered the reverse of if..., then... rather than the same
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– Henry
Nov 13 '18 at 8:46
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@Henry - reverse ?
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– Mauro ALLEGRANZA
Nov 13 '18 at 9:16
2
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What does the first word (exactly) refer to?
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– Carsten S
Nov 13 '18 at 10:40
add a comment |
$begingroup$
Why isn’t ‘because’ a logical connective in propositional logic?
Is this because the equivalent of ‘because’ is the argument of the form ‘if $p$, then $q$’ ?
Exactly.
Either the connective "because" is truth-functional, in which case it is the same as "if..., then...", or it is not truth-functional, in which case we need a different way of modelling it.
See e.g. Counterfactual Theories of Causation.
See also Arthur Burks, The Logic of Causal Proposition, Mind (1951).
$endgroup$
$begingroup$
I would have thought a truth-functional because might be considered the reverse of if..., then... rather than the same
$endgroup$
– Henry
Nov 13 '18 at 8:46
$begingroup$
@Henry - reverse ?
$endgroup$
– Mauro ALLEGRANZA
Nov 13 '18 at 9:16
2
$begingroup$
What does the first word (exactly) refer to?
$endgroup$
– Carsten S
Nov 13 '18 at 10:40
add a comment |
$begingroup$
Why isn’t ‘because’ a logical connective in propositional logic?
Is this because the equivalent of ‘because’ is the argument of the form ‘if $p$, then $q$’ ?
Exactly.
Either the connective "because" is truth-functional, in which case it is the same as "if..., then...", or it is not truth-functional, in which case we need a different way of modelling it.
See e.g. Counterfactual Theories of Causation.
See also Arthur Burks, The Logic of Causal Proposition, Mind (1951).
$endgroup$
Why isn’t ‘because’ a logical connective in propositional logic?
Is this because the equivalent of ‘because’ is the argument of the form ‘if $p$, then $q$’ ?
Exactly.
Either the connective "because" is truth-functional, in which case it is the same as "if..., then...", or it is not truth-functional, in which case we need a different way of modelling it.
See e.g. Counterfactual Theories of Causation.
See also Arthur Burks, The Logic of Causal Proposition, Mind (1951).
edited Nov 13 '18 at 10:49
answered Nov 11 '18 at 13:17
Mauro ALLEGRANZAMauro ALLEGRANZA
65.8k449114
65.8k449114
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I would have thought a truth-functional because might be considered the reverse of if..., then... rather than the same
$endgroup$
– Henry
Nov 13 '18 at 8:46
$begingroup$
@Henry - reverse ?
$endgroup$
– Mauro ALLEGRANZA
Nov 13 '18 at 9:16
2
$begingroup$
What does the first word (exactly) refer to?
$endgroup$
– Carsten S
Nov 13 '18 at 10:40
add a comment |
$begingroup$
I would have thought a truth-functional because might be considered the reverse of if..., then... rather than the same
$endgroup$
– Henry
Nov 13 '18 at 8:46
$begingroup$
@Henry - reverse ?
$endgroup$
– Mauro ALLEGRANZA
Nov 13 '18 at 9:16
2
$begingroup$
What does the first word (exactly) refer to?
$endgroup$
– Carsten S
Nov 13 '18 at 10:40
$begingroup$
I would have thought a truth-functional because might be considered the reverse of if..., then... rather than the same
$endgroup$
– Henry
Nov 13 '18 at 8:46
$begingroup$
I would have thought a truth-functional because might be considered the reverse of if..., then... rather than the same
$endgroup$
– Henry
Nov 13 '18 at 8:46
$begingroup$
@Henry - reverse ?
$endgroup$
– Mauro ALLEGRANZA
Nov 13 '18 at 9:16
$begingroup$
@Henry - reverse ?
$endgroup$
– Mauro ALLEGRANZA
Nov 13 '18 at 9:16
2
2
$begingroup$
What does the first word (exactly) refer to?
$endgroup$
– Carsten S
Nov 13 '18 at 10:40
$begingroup$
What does the first word (exactly) refer to?
$endgroup$
– Carsten S
Nov 13 '18 at 10:40
add a comment |
$begingroup$
I agree with the other answers, however I want to add that the closest thing might be the turnstyle symbol $vdash$, although this is usually read as "yields", and thus points the other way. If I write
$$A vdash B$$
this is read as "A yields B", or "knowing A, I can prove B". If you wanted to encode because, you could probably read it backwards as "B because of A".
Note however that this is not used as part of a logical formula, but as a shorthand between formulas when writing down a proof. So $A vdash B$ is no longer a formula, but rather a statement on how to prove $B$. (In most of the rest of mathematics, you would write $Rightarrow$ in your proof instead, however in logic this is of course easily confused with the implication inside formulas)
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2
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Yes, this would therefore be a 'meta-logical' symbol;: a symbol used to say something about logic exprssion .. but it is not a logical connective or operator.
$endgroup$
– Bram28
Nov 12 '18 at 12:21
add a comment |
$begingroup$
I agree with the other answers, however I want to add that the closest thing might be the turnstyle symbol $vdash$, although this is usually read as "yields", and thus points the other way. If I write
$$A vdash B$$
this is read as "A yields B", or "knowing A, I can prove B". If you wanted to encode because, you could probably read it backwards as "B because of A".
Note however that this is not used as part of a logical formula, but as a shorthand between formulas when writing down a proof. So $A vdash B$ is no longer a formula, but rather a statement on how to prove $B$. (In most of the rest of mathematics, you would write $Rightarrow$ in your proof instead, however in logic this is of course easily confused with the implication inside formulas)
$endgroup$
2
$begingroup$
Yes, this would therefore be a 'meta-logical' symbol;: a symbol used to say something about logic exprssion .. but it is not a logical connective or operator.
$endgroup$
– Bram28
Nov 12 '18 at 12:21
add a comment |
$begingroup$
I agree with the other answers, however I want to add that the closest thing might be the turnstyle symbol $vdash$, although this is usually read as "yields", and thus points the other way. If I write
$$A vdash B$$
this is read as "A yields B", or "knowing A, I can prove B". If you wanted to encode because, you could probably read it backwards as "B because of A".
Note however that this is not used as part of a logical formula, but as a shorthand between formulas when writing down a proof. So $A vdash B$ is no longer a formula, but rather a statement on how to prove $B$. (In most of the rest of mathematics, you would write $Rightarrow$ in your proof instead, however in logic this is of course easily confused with the implication inside formulas)
$endgroup$
I agree with the other answers, however I want to add that the closest thing might be the turnstyle symbol $vdash$, although this is usually read as "yields", and thus points the other way. If I write
$$A vdash B$$
this is read as "A yields B", or "knowing A, I can prove B". If you wanted to encode because, you could probably read it backwards as "B because of A".
Note however that this is not used as part of a logical formula, but as a shorthand between formulas when writing down a proof. So $A vdash B$ is no longer a formula, but rather a statement on how to prove $B$. (In most of the rest of mathematics, you would write $Rightarrow$ in your proof instead, however in logic this is of course easily confused with the implication inside formulas)
answered Nov 11 '18 at 20:32
mlkmlk
2,8851016
2,8851016
2
$begingroup$
Yes, this would therefore be a 'meta-logical' symbol;: a symbol used to say something about logic exprssion .. but it is not a logical connective or operator.
$endgroup$
– Bram28
Nov 12 '18 at 12:21
add a comment |
2
$begingroup$
Yes, this would therefore be a 'meta-logical' symbol;: a symbol used to say something about logic exprssion .. but it is not a logical connective or operator.
$endgroup$
– Bram28
Nov 12 '18 at 12:21
2
2
$begingroup$
Yes, this would therefore be a 'meta-logical' symbol;: a symbol used to say something about logic exprssion .. but it is not a logical connective or operator.
$endgroup$
– Bram28
Nov 12 '18 at 12:21
$begingroup$
Yes, this would therefore be a 'meta-logical' symbol;: a symbol used to say something about logic exprssion .. but it is not a logical connective or operator.
$endgroup$
– Bram28
Nov 12 '18 at 12:21
add a comment |
$begingroup$
You can define things however you want. (Be careful; you may accidentally be inconsistent.)
Either:
- "because" is logically equivalent to a binary operator
- or it's not
If it is, it's probably the same as "only if" (Or take your pick of the other 15 operators). Adding a "because" overload would create an additional word to remember: unneeded complexity. We like simplicity.
If it isn't, you can define a binary function because(a, b)
however you want.
$endgroup$
add a comment |
$begingroup$
You can define things however you want. (Be careful; you may accidentally be inconsistent.)
Either:
- "because" is logically equivalent to a binary operator
- or it's not
If it is, it's probably the same as "only if" (Or take your pick of the other 15 operators). Adding a "because" overload would create an additional word to remember: unneeded complexity. We like simplicity.
If it isn't, you can define a binary function because(a, b)
however you want.
$endgroup$
add a comment |
$begingroup$
You can define things however you want. (Be careful; you may accidentally be inconsistent.)
Either:
- "because" is logically equivalent to a binary operator
- or it's not
If it is, it's probably the same as "only if" (Or take your pick of the other 15 operators). Adding a "because" overload would create an additional word to remember: unneeded complexity. We like simplicity.
If it isn't, you can define a binary function because(a, b)
however you want.
$endgroup$
You can define things however you want. (Be careful; you may accidentally be inconsistent.)
Either:
- "because" is logically equivalent to a binary operator
- or it's not
If it is, it's probably the same as "only if" (Or take your pick of the other 15 operators). Adding a "because" overload would create an additional word to remember: unneeded complexity. We like simplicity.
If it isn't, you can define a binary function because(a, b)
however you want.
answered Nov 13 '18 at 5:32
Words Like JaredWords Like Jared
1337
1337
add a comment |
add a comment |
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"because" is about causality, not implication
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– Hagen von Eitzen
Nov 11 '18 at 13:17
1
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Folks might find basicbooks.com/titles/judea-pearl/the-book-of-why/9780465097609 of interest.
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– Barry Cipra
Nov 11 '18 at 13:23
1
$begingroup$
I would love to see a complete prepositional logic theory where P→Q is not equivalent to ¬P∨Q but I haven't yet. :(
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– Joshua
Nov 12 '18 at 19:10
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@Joshua Maybe I'm misunderstanding what you mean, but that equivalence typically fails in non-classical logics. It definitely fails in constructive/intuitionistic logic. So Intuitionistic Propositional Logic would be a propositional logic where $Pto Q$ is not equivalent to $neg Plor Q$.
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– Derek Elkins
Nov 13 '18 at 6:23
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$(Pto Q) lor (Qto P)$ is a tautology in classical logic, so presumably $(Qleftarrow P) lor (Pleftarrow Q)$ is too. You would not want to read "$leftarrow$" as because here
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– Henry
Nov 13 '18 at 8:41