Mixing
Is determining derivability in propositional logic decidable?
$begingroup$
I know that determining (semantic) entailment in propositional logic is decidable by the truth table method. For instance, let:
$phi = b rightarrow a \ psi = b lor c rightarrow a$
Then we can use the truth table method to determine whether $phi$ entails $sigma$, which is this case is true.
We can also (syntactically) derive $psi$ from $phi$, for instance by using natural deduction. However, is determining derivability decidable in propositional logic?
logic propositional-calculus proof-theory decidability
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
$endgroup$
add a comment |
$begingroup$
I know that determining (semantic) entailment in propositional logic is decidable by the truth table method. For instance, let:
$phi = b rightarrow a \ psi = b lor c rightarrow a$
Then we can use the truth table method to determine whether $phi$ entails $sigma$, which is this case is true.
We can also (syntactically) derive $psi$ from $phi$, for instance by using natural deduction. However, is determining derivability decidable in propositional logic?
logic propositional-calculus proof-theory decidability
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
$endgroup$
$begingroup$
Yes by the completeness theorem. If $Gamma $ entails $phi $ and $Gamma $ is finite, then $bigwedgeGammaimpliesphi $ is valid, it is derivable by the completeness theorem, $phi $ is derivable from $Gamma $ by modus ponens. The soundness theorem is used in the converse direction.
$endgroup$
– beroal
Sep 25 '17 at 16:40
$begingroup$
I thought the completeness theorem stated that there are rules of inference which are sufficient to derive any entailed proposition, i.e. if $phi models psi$ then $phi rightarrow psi$. I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable, just as in the case of first-order logic.
$endgroup$
– S0rin
Sep 26 '17 at 8:45
$begingroup$
"I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable" This is a consequence. There is a decision procedure for entailment, and entailment is equivalent to derivability.
$endgroup$
– beroal
Sep 27 '17 at 5:43
$begingroup$
Even if we didn’t know about completeness and semantics, there are direct ways to see that provability is decidable, e.g. cut elimination. Sometimes this can be the easiest way to show it in more exotic logics where it’s harder to show the semantics are decidable.
$endgroup$
– spaceisdarkgreen
Jan 21 at 22:33
add a comment |
$begingroup$
I know that determining (semantic) entailment in propositional logic is decidable by the truth table method. For instance, let:
$phi = b rightarrow a \ psi = b lor c rightarrow a$
Then we can use the truth table method to determine whether $phi$ entails $sigma$, which is this case is true.
We can also (syntactically) derive $psi$ from $phi$, for instance by using natural deduction. However, is determining derivability decidable in propositional logic?
logic propositional-calculus proof-theory decidability
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
$endgroup$
I know that determining (semantic) entailment in propositional logic is decidable by the truth table method. For instance, let:
$phi = b rightarrow a \ psi = b lor c rightarrow a$
Then we can use the truth table method to determine whether $phi$ entails $sigma$, which is this case is true.
We can also (syntactically) derive $psi$ from $phi$, for instance by using natural deduction. However, is determining derivability decidable in propositional logic?
logic propositional-calculus proof-theory decidability
logic propositional-calculus proof-theory decidability
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
edited Jan 21 at 21:50
Taroccoesbrocco
5,64271840
5,64271840
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
asked Sep 25 '17 at 13:34
S0rinS0rin
25119
25119
$begingroup$
Yes by the completeness theorem. If $Gamma $ entails $phi $ and $Gamma $ is finite, then $bigwedgeGammaimpliesphi $ is valid, it is derivable by the completeness theorem, $phi $ is derivable from $Gamma $ by modus ponens. The soundness theorem is used in the converse direction.
$endgroup$
– beroal
Sep 25 '17 at 16:40
$begingroup$
I thought the completeness theorem stated that there are rules of inference which are sufficient to derive any entailed proposition, i.e. if $phi models psi$ then $phi rightarrow psi$. I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable, just as in the case of first-order logic.
$endgroup$
– S0rin
Sep 26 '17 at 8:45
$begingroup$
"I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable" This is a consequence. There is a decision procedure for entailment, and entailment is equivalent to derivability.
$endgroup$
– beroal
Sep 27 '17 at 5:43
$begingroup$
Even if we didn’t know about completeness and semantics, there are direct ways to see that provability is decidable, e.g. cut elimination. Sometimes this can be the easiest way to show it in more exotic logics where it’s harder to show the semantics are decidable.
$endgroup$
– spaceisdarkgreen
Jan 21 at 22:33
add a comment |
$begingroup$
Yes by the completeness theorem. If $Gamma $ entails $phi $ and $Gamma $ is finite, then $bigwedgeGammaimpliesphi $ is valid, it is derivable by the completeness theorem, $phi $ is derivable from $Gamma $ by modus ponens. The soundness theorem is used in the converse direction.
$endgroup$
– beroal
Sep 25 '17 at 16:40
$begingroup$
I thought the completeness theorem stated that there are rules of inference which are sufficient to derive any entailed proposition, i.e. if $phi models psi$ then $phi rightarrow psi$. I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable, just as in the case of first-order logic.
$endgroup$
– S0rin
Sep 26 '17 at 8:45
$begingroup$
"I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable" This is a consequence. There is a decision procedure for entailment, and entailment is equivalent to derivability.
$endgroup$
– beroal
Sep 27 '17 at 5:43
$begingroup$
Even if we didn’t know about completeness and semantics, there are direct ways to see that provability is decidable, e.g. cut elimination. Sometimes this can be the easiest way to show it in more exotic logics where it’s harder to show the semantics are decidable.
$endgroup$
– spaceisdarkgreen
Jan 21 at 22:33
$begingroup$
Yes by the completeness theorem. If $Gamma $ entails $phi $ and $Gamma $ is finite, then $bigwedgeGammaimpliesphi $ is valid, it is derivable by the completeness theorem, $phi $ is derivable from $Gamma $ by modus ponens. The soundness theorem is used in the converse direction.
$endgroup$
– beroal
Sep 25 '17 at 16:40
$begingroup$
Yes by the completeness theorem. If $Gamma $ entails $phi $ and $Gamma $ is finite, then $bigwedgeGammaimpliesphi $ is valid, it is derivable by the completeness theorem, $phi $ is derivable from $Gamma $ by modus ponens. The soundness theorem is used in the converse direction.
$endgroup$
– beroal
Sep 25 '17 at 16:40
$begingroup$
I thought the completeness theorem stated that there are rules of inference which are sufficient to derive any entailed proposition, i.e. if $phi models psi$ then $phi rightarrow psi$. I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable, just as in the case of first-order logic.
$endgroup$
– S0rin
Sep 26 '17 at 8:45
$begingroup$
I thought the completeness theorem stated that there are rules of inference which are sufficient to derive any entailed proposition, i.e. if $phi models psi$ then $phi rightarrow psi$. I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable, just as in the case of first-order logic.
$endgroup$
– S0rin
Sep 26 '17 at 8:45
$begingroup$
"I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable" This is a consequence. There is a decision procedure for entailment, and entailment is equivalent to derivability.
$endgroup$
– beroal
Sep 27 '17 at 5:43
$begingroup$
"I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable" This is a consequence. There is a decision procedure for entailment, and entailment is equivalent to derivability.
$endgroup$
– beroal
Sep 27 '17 at 5:43
$begingroup$
Even if we didn’t know about completeness and semantics, there are direct ways to see that provability is decidable, e.g. cut elimination. Sometimes this can be the easiest way to show it in more exotic logics where it’s harder to show the semantics are decidable.
$endgroup$
– spaceisdarkgreen
Jan 21 at 22:33
$begingroup$
Even if we didn’t know about completeness and semantics, there are direct ways to see that provability is decidable, e.g. cut elimination. Sometimes this can be the easiest way to show it in more exotic logics where it’s harder to show the semantics are decidable.
$endgroup$
– spaceisdarkgreen
Jan 21 at 22:33
add a comment |
1 Answer
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$begingroup$
For systems that are both sound and complete (which for propositional logic is pretty much any system, that's out there ... it would be embarrassing for a logician to propose a proof system for propositional logic that would be not be both sound and complete), this is trivial: we would have $Gamma vdash phi$ iff $Gamma vDash phi$, and since the latter is decidable, the former is decidable as well.
If a proof system is not both sound and complete .... well, if a system is not sound and is in fact able to derive any statement from nothing, then derivability is also trivially decidable: everything is derivable!
For other systems yet ... Hmmm, I am actually unsure of the answer ... though I wonder if you could have some kind of bizarre propositional logic with which you can encode claims about Turing-machines and their behavior, and thus generate some kind of undecidability result ...
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
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$begingroup$
For systems that are both sound and complete (which for propositional logic is pretty much any system, that's out there ... it would be embarrassing for a logician to propose a proof system for propositional logic that would be not be both sound and complete), this is trivial: we would have $Gamma vdash phi$ iff $Gamma vDash phi$, and since the latter is decidable, the former is decidable as well.
If a proof system is not both sound and complete .... well, if a system is not sound and is in fact able to derive any statement from nothing, then derivability is also trivially decidable: everything is derivable!
For other systems yet ... Hmmm, I am actually unsure of the answer ... though I wonder if you could have some kind of bizarre propositional logic with which you can encode claims about Turing-machines and their behavior, and thus generate some kind of undecidability result ...
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
$begingroup$
For systems that are both sound and complete (which for propositional logic is pretty much any system, that's out there ... it would be embarrassing for a logician to propose a proof system for propositional logic that would be not be both sound and complete), this is trivial: we would have $Gamma vdash phi$ iff $Gamma vDash phi$, and since the latter is decidable, the former is decidable as well.
If a proof system is not both sound and complete .... well, if a system is not sound and is in fact able to derive any statement from nothing, then derivability is also trivially decidable: everything is derivable!
For other systems yet ... Hmmm, I am actually unsure of the answer ... though I wonder if you could have some kind of bizarre propositional logic with which you can encode claims about Turing-machines and their behavior, and thus generate some kind of undecidability result ...
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
$endgroup$
add a comment |
$begingroup$
For systems that are both sound and complete (which for propositional logic is pretty much any system, that's out there ... it would be embarrassing for a logician to propose a proof system for propositional logic that would be not be both sound and complete), this is trivial: we would have $Gamma vdash phi$ iff $Gamma vDash phi$, and since the latter is decidable, the former is decidable as well.
If a proof system is not both sound and complete .... well, if a system is not sound and is in fact able to derive any statement from nothing, then derivability is also trivially decidable: everything is derivable!
For other systems yet ... Hmmm, I am actually unsure of the answer ... though I wonder if you could have some kind of bizarre propositional logic with which you can encode claims about Turing-machines and their behavior, and thus generate some kind of undecidability result ...
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
$endgroup$
For systems that are both sound and complete (which for propositional logic is pretty much any system, that's out there ... it would be embarrassing for a logician to propose a proof system for propositional logic that would be not be both sound and complete), this is trivial: we would have $Gamma vdash phi$ iff $Gamma vDash phi$, and since the latter is decidable, the former is decidable as well.
If a proof system is not both sound and complete .... well, if a system is not sound and is in fact able to derive any statement from nothing, then derivability is also trivially decidable: everything is derivable!
For other systems yet ... Hmmm, I am actually unsure of the answer ... though I wonder if you could have some kind of bizarre propositional logic with which you can encode claims about Turing-machines and their behavior, and thus generate some kind of undecidability result ...
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
edited Sep 25 '17 at 17:41
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
answered Sep 25 '17 at 16:56
Bram28Bram28
63.3k44793
63.3k44793
add a comment |
add a comment |
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$begingroup$
Yes by the completeness theorem. If $Gamma $ entails $phi $ and $Gamma $ is finite, then $bigwedgeGammaimpliesphi $ is valid, it is derivable by the completeness theorem, $phi $ is derivable from $Gamma $ by modus ponens. The soundness theorem is used in the converse direction.
$endgroup$
– beroal
Sep 25 '17 at 16:40
$begingroup$
I thought the completeness theorem stated that there are rules of inference which are sufficient to derive any entailed proposition, i.e. if $phi models psi$ then $phi rightarrow psi$. I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable, just as in the case of first-order logic.
$endgroup$
– S0rin
Sep 26 '17 at 8:45
$begingroup$
"I didn't think this theorem implied that there is a decision procedure to determine whether a proposition is derivable" This is a consequence. There is a decision procedure for entailment, and entailment is equivalent to derivability.
$endgroup$
– beroal
Sep 27 '17 at 5:43
$begingroup$
Even if we didn’t know about completeness and semantics, there are direct ways to see that provability is decidable, e.g. cut elimination. Sometimes this can be the easiest way to show it in more exotic logics where it’s harder to show the semantics are decidable.
$endgroup$
– spaceisdarkgreen
Jan 21 at 22:33