What does a cdot/interpunct stand for in a (logical) set notation?
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I'm studying Computation Tree Logic for a course at university. However, I seem to have forgotten an important detail about set builder notation that I haven't found online yet. I'm wondering if this is just a creative notation used by my university.
The formulas I encounter describe formal semantics of CTL in the form of:
$M ,s models p$ iff $p;epsilon ; L(s)$
However, some are more complex, including an interpunct:
$M ,s models AX f$ iff $forallpi;epsilon ; Pi(M,s) cdot M,pi[1] models f$
And below, which defines the set of execution paths for a Kripke structure:
$Pi(M,s) equiv $ { $pi;|;pi[0] = s wedgeforall n ;cdot (pi[n], pi[n+1]);epsilon;R$ }
What does the $cdot$ mean here?
logic notation
asked Nov 22 at 14:38
Zimano
1276
|
show 3 more comments
up vote
1
down vote
favorite
I'm studying Computation Tree Logic for a course at university. However, I seem to have forgotten an important detail about set builder notation that I haven't found online yet. I'm wondering if this is just a creative notation used by my university.
The formulas I encounter describe formal semantics of CTL in the form of:
$M ,s models p$ iff $p;epsilon ; L(s)$
However, some are more complex, including an interpunct:
$M ,s models AX f$ iff $forallpi;epsilon ; Pi(M,s) cdot M,pi[1] models f$
And below, which defines the set of execution paths for a Kripke structure:
$Pi(M,s) equiv $ { $pi;|;pi[0] = s wedgeforall n ;cdot (pi[n], pi[n+1]);epsilon;R$ }
What does the $cdot$ mean here?
logic notation
asked Nov 22 at 14:38
Zimano
1276
Nothing. It says $M,s vDash AXf$ holds iff "for every $pi$ in $Pi(M,s)$ we have that $M, pi[1] vDash f$ holds".
– Mauro ALLEGRANZA
Nov 22 at 14:41
@MauroALLEGRANZA If it meant nothing then could it be omitted without changing the formula? Why aren't comma's used then?
– Zimano
Nov 22 at 14:41
It is not set-builder notation.
– Mauro ALLEGRANZA
Nov 22 at 14:43
@MauroALLEGRANZA That's another possibility. I'll update my question to include a case where set builder notation was used together with an interpunct. The examples I gave are indeed not written as such.
– Zimano
Nov 22 at 14:44
Oldish logic texts used notation where periods played a role similar to what we now do with parentheses. This seems to be a remnant.
– Andrés E. Caicedo
Nov 22 at 14:51
|
show 3 more comments
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I'm studying Computation Tree Logic for a course at university. However, I seem to have forgotten an important detail about set builder notation that I haven't found online yet. I'm wondering if this is just a creative notation used by my university.
The formulas I encounter describe formal semantics of CTL in the form of:
$M ,s models p$ iff $p;epsilon ; L(s)$
However, some are more complex, including an interpunct:
$M ,s models AX f$ iff $forallpi;epsilon ; Pi(M,s) cdot M,pi[1] models f$
And below, which defines the set of execution paths for a Kripke structure:
$Pi(M,s) equiv $ { $pi;|;pi[0] = s wedgeforall n ;cdot (pi[n], pi[n+1]);epsilon;R$ }
What does the $cdot$ mean here?
logic notation
asked Nov 22 at 14:38
Zimano
1276
I'm studying Computation Tree Logic for a course at university. However, I seem to have forgotten an important detail about set builder notation that I haven't found online yet. I'm wondering if this is just a creative notation used by my university.
The formulas I encounter describe formal semantics of CTL in the form of:
$M ,s models p$ iff $p;epsilon ; L(s)$
However, some are more complex, including an interpunct:
$M ,s models AX f$ iff $forallpi;epsilon ; Pi(M,s) cdot M,pi[1] models f$
And below, which defines the set of execution paths for a Kripke structure:
$Pi(M,s) equiv $ { $pi;|;pi[0] = s wedgeforall n ;cdot (pi[n], pi[n+1]);epsilon;R$ }
What does the $cdot$ mean here?
logic notation
logic notation
asked Nov 22 at 14:38
Zimano
1276
asked Nov 22 at 14:38
Zimano
1276
edited Nov 22 at 14:48
asked Nov 22 at 14:38
Zimano
1276
asked Nov 22 at 14:38
Zimano
1276
asked Nov 22 at 14:38
Zimano
1276
1276
Nothing. It says $M,s vDash AXf$ holds iff "for every $pi$ in $Pi(M,s)$ we have that $M, pi[1] vDash f$ holds".
– Mauro ALLEGRANZA
Nov 22 at 14:41
@MauroALLEGRANZA If it meant nothing then could it be omitted without changing the formula? Why aren't comma's used then?
– Zimano
Nov 22 at 14:41
It is not set-builder notation.
– Mauro ALLEGRANZA
Nov 22 at 14:43
@MauroALLEGRANZA That's another possibility. I'll update my question to include a case where set builder notation was used together with an interpunct. The examples I gave are indeed not written as such.
– Zimano
Nov 22 at 14:44
Oldish logic texts used notation where periods played a role similar to what we now do with parentheses. This seems to be a remnant.
– Andrés E. Caicedo
Nov 22 at 14:51
|
show 3 more comments
Nothing. It says $M,s vDash AXf$ holds iff "for every $pi$ in $Pi(M,s)$ we have that $M, pi[1] vDash f$ holds".
– Mauro ALLEGRANZA
Nov 22 at 14:41
@MauroALLEGRANZA If it meant nothing then could it be omitted without changing the formula? Why aren't comma's used then?
– Zimano
Nov 22 at 14:41
It is not set-builder notation.
– Mauro ALLEGRANZA
Nov 22 at 14:43
@MauroALLEGRANZA That's another possibility. I'll update my question to include a case where set builder notation was used together with an interpunct. The examples I gave are indeed not written as such.
– Zimano
Nov 22 at 14:44
Oldish logic texts used notation where periods played a role similar to what we now do with parentheses. This seems to be a remnant.
– Andrés E. Caicedo
Nov 22 at 14:51
Nothing. It says $M,s vDash AXf$ holds iff "for every $pi$ in $Pi(M,s)$ we have that $M, pi[1] vDash f$ holds".
– Mauro ALLEGRANZA
Nov 22 at 14:41
Nothing. It says $M,s vDash AXf$ holds iff "for every $pi$ in $Pi(M,s)$ we have that $M, pi[1] vDash f$ holds".
– Mauro ALLEGRANZA
Nov 22 at 14:41
@MauroALLEGRANZA If it meant nothing then could it be omitted without changing the formula? Why aren't comma's used then?
– Zimano
Nov 22 at 14:41
@MauroALLEGRANZA If it meant nothing then could it be omitted without changing the formula? Why aren't comma's used then?
– Zimano
Nov 22 at 14:41
It is not set-builder notation.
– Mauro ALLEGRANZA
Nov 22 at 14:43
It is not set-builder notation.
– Mauro ALLEGRANZA
Nov 22 at 14:43
@MauroALLEGRANZA That's another possibility. I'll update my question to include a case where set builder notation was used together with an interpunct. The examples I gave are indeed not written as such.
– Zimano
Nov 22 at 14:44
@MauroALLEGRANZA That's another possibility. I'll update my question to include a case where set builder notation was used together with an interpunct. The examples I gave are indeed not written as such.
– Zimano
Nov 22 at 14:44
Oldish logic texts used notation where periods played a role similar to what we now do with parentheses. This seems to be a remnant.
– Andrés E. Caicedo
Nov 22 at 14:51
Oldish logic texts used notation where periods played a role similar to what we now do with parentheses. This seems to be a remnant.
– Andrés E. Caicedo
Nov 22 at 14:51
|
show 3 more comments
1 Answer
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oldest
votes
up vote
1
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It looks like the dot is just part of the author's notation for quantifiers:
$$ forall x in A cdot varphi(x) $$
means "the property $varphi$ holds for all elements of the set $A$". The dot is just there to separate the $A$ from the $varphi$ (since each of them can be multiple symbols long).
In $forall n cdot (pi[n],pi[n+1]in R$ there is no $in A$ part, but the dot is still there to separate the quantifier from the formula being quantified over.
answered Nov 22 at 14:51
Henning Makholm
236k16300534
Thanks! I'm starting to understand the usage now looking at the other material. It was pretty evident, but my brain just gooped all over it.
– Zimano
Nov 22 at 14:54
add a comment |
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It looks like the dot is just part of the author's notation for quantifiers:
$$ forall x in A cdot varphi(x) $$
means "the property $varphi$ holds for all elements of the set $A$". The dot is just there to separate the $A$ from the $varphi$ (since each of them can be multiple symbols long).
In $forall n cdot (pi[n],pi[n+1]in R$ there is no $in A$ part, but the dot is still there to separate the quantifier from the formula being quantified over.
answered Nov 22 at 14:51
Henning Makholm
236k16300534
Thanks! I'm starting to understand the usage now looking at the other material. It was pretty evident, but my brain just gooped all over it.
– Zimano
Nov 22 at 14:54
add a comment |
up vote
1
down vote
accepted
It looks like the dot is just part of the author's notation for quantifiers:
$$ forall x in A cdot varphi(x) $$
means "the property $varphi$ holds for all elements of the set $A$". The dot is just there to separate the $A$ from the $varphi$ (since each of them can be multiple symbols long).
In $forall n cdot (pi[n],pi[n+1]in R$ there is no $in A$ part, but the dot is still there to separate the quantifier from the formula being quantified over.
answered Nov 22 at 14:51
Henning Makholm
236k16300534
Thanks! I'm starting to understand the usage now looking at the other material. It was pretty evident, but my brain just gooped all over it.
– Zimano
Nov 22 at 14:54
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It looks like the dot is just part of the author's notation for quantifiers:
$$ forall x in A cdot varphi(x) $$
means "the property $varphi$ holds for all elements of the set $A$". The dot is just there to separate the $A$ from the $varphi$ (since each of them can be multiple symbols long).
In $forall n cdot (pi[n],pi[n+1]in R$ there is no $in A$ part, but the dot is still there to separate the quantifier from the formula being quantified over.
answered Nov 22 at 14:51
Henning Makholm
236k16300534
It looks like the dot is just part of the author's notation for quantifiers:
$$ forall x in A cdot varphi(x) $$
means "the property $varphi$ holds for all elements of the set $A$". The dot is just there to separate the $A$ from the $varphi$ (since each of them can be multiple symbols long).
In $forall n cdot (pi[n],pi[n+1]in R$ there is no $in A$ part, but the dot is still there to separate the quantifier from the formula being quantified over.
answered Nov 22 at 14:51
Henning Makholm
236k16300534
answered Nov 22 at 14:51
Henning Makholm
236k16300534
answered Nov 22 at 14:51
Henning Makholm
236k16300534
answered Nov 22 at 14:51
Henning Makholm
236k16300534
236k16300534
Thanks! I'm starting to understand the usage now looking at the other material. It was pretty evident, but my brain just gooped all over it.
– Zimano
Nov 22 at 14:54
add a comment |
Thanks! I'm starting to understand the usage now looking at the other material. It was pretty evident, but my brain just gooped all over it.
– Zimano
Nov 22 at 14:54
Thanks! I'm starting to understand the usage now looking at the other material. It was pretty evident, but my brain just gooped all over it.
– Zimano
Nov 22 at 14:54
Thanks! I'm starting to understand the usage now looking at the other material. It was pretty evident, but my brain just gooped all over it.
– Zimano
Nov 22 at 14:54
add a comment |
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Nothing. It says $M,s vDash AXf$ holds iff "for every $pi$ in $Pi(M,s)$ we have that $M, pi[1] vDash f$ holds".
– Mauro ALLEGRANZA
Nov 22 at 14:41
@MauroALLEGRANZA If it meant nothing then could it be omitted without changing the formula? Why aren't comma's used then?
– Zimano
Nov 22 at 14:41
It is not set-builder notation.
– Mauro ALLEGRANZA
Nov 22 at 14:43
@MauroALLEGRANZA That's another possibility. I'll update my question to include a case where set builder notation was used together with an interpunct. The examples I gave are indeed not written as such.
– Zimano
Nov 22 at 14:44
Oldish logic texts used notation where periods played a role similar to what we now do with parentheses. This seems to be a remnant.
– Andrés E. Caicedo
Nov 22 at 14:51