Mixing
Negation - Some operating systems always crash
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I am trying to find the negation of the statement "Some operating systems always crash"
I know that the negation of "some" is "all" so:
All operation systems always crash ?
Or:
All operation systems never crash ?
I don't understand what to do with the "always" in this statement.
Does anyone know the answer to this?
discrete-mathematics
asked Feb 1 at 13:30
CUPACUPA
395
$endgroup$
add a comment |
$begingroup$
I am trying to find the negation of the statement "Some operating systems always crash"
I know that the negation of "some" is "all" so:
All operation systems always crash ?
Or:
All operation systems never crash ?
I don't understand what to do with the "always" in this statement.
Does anyone know the answer to this?
discrete-mathematics
asked Feb 1 at 13:30
CUPACUPA
395
$endgroup$
2
$begingroup$
The negation of "some" is not "all". "Some" means "at least one". Thus, the negation of "some" is "not some", i.e. "all not".
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:33
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"always" is tricky here; if you have not a specific need to express the temporal fact, we can symply say : "Some operating systems always-crash", i.e. $exists x (text {OpSys}(x) land text {Crash}(x))$.
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:38
add a comment |
$begingroup$
I am trying to find the negation of the statement "Some operating systems always crash"
I know that the negation of "some" is "all" so:
All operation systems always crash ?
Or:
All operation systems never crash ?
I don't understand what to do with the "always" in this statement.
Does anyone know the answer to this?
discrete-mathematics
asked Feb 1 at 13:30
CUPACUPA
395
$endgroup$
I am trying to find the negation of the statement "Some operating systems always crash"
I know that the negation of "some" is "all" so:
All operation systems always crash ?
Or:
All operation systems never crash ?
I don't understand what to do with the "always" in this statement.
Does anyone know the answer to this?
discrete-mathematics
discrete-mathematics
asked Feb 1 at 13:30
CUPACUPA
395
asked Feb 1 at 13:30
CUPACUPA
395
asked Feb 1 at 13:30
CUPACUPA
395
asked Feb 1 at 13:30
CUPACUPA
395
asked Feb 1 at 13:30
CUPACUPA
395
395
2
$begingroup$
The negation of "some" is not "all". "Some" means "at least one". Thus, the negation of "some" is "not some", i.e. "all not".
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:33
$begingroup$
"always" is tricky here; if you have not a specific need to express the temporal fact, we can symply say : "Some operating systems always-crash", i.e. $exists x (text {OpSys}(x) land text {Crash}(x))$.
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:38
add a comment |
2
$begingroup$
The negation of "some" is not "all". "Some" means "at least one". Thus, the negation of "some" is "not some", i.e. "all not".
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:33
$begingroup$
"always" is tricky here; if you have not a specific need to express the temporal fact, we can symply say : "Some operating systems always-crash", i.e. $exists x (text {OpSys}(x) land text {Crash}(x))$.
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:38
2
2
$begingroup$
The negation of "some" is not "all". "Some" means "at least one". Thus, the negation of "some" is "not some", i.e. "all not".
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:33
$begingroup$
The negation of "some" is not "all". "Some" means "at least one". Thus, the negation of "some" is "not some", i.e. "all not".
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:33
$begingroup$
"always" is tricky here; if you have not a specific need to express the temporal fact, we can symply say : "Some operating systems always-crash", i.e. $exists x (text {OpSys}(x) land text {Crash}(x))$.
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:38
$begingroup$
"always" is tricky here; if you have not a specific need to express the temporal fact, we can symply say : "Some operating systems always-crash", i.e. $exists x (text {OpSys}(x) land text {Crash}(x))$.
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:38
add a comment |
1 Answer
1
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oldest
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$begingroup$
One way to look at the statement "some operating systems always crash":
$exists O in textrm{OS}: forall t: c(O,t)$, where I use OS for the set of operating systems and $t$ quantifies over times, $c(O,t)$ means the OS $O$ crashes at time $t$.
If you agree that this is the intended translation (this is a matter of linguistic discussion) then its logical negation is
$forall O in textrm{OS}: exists t: lnot c(O,t)$, by the usual rules for first order logic, which in English I would translate as "every operating system sometimes doesn't crash".
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
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add a comment |
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1 Answer
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1 Answer
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$begingroup$
One way to look at the statement "some operating systems always crash":
$exists O in textrm{OS}: forall t: c(O,t)$, where I use OS for the set of operating systems and $t$ quantifies over times, $c(O,t)$ means the OS $O$ crashes at time $t$.
If you agree that this is the intended translation (this is a matter of linguistic discussion) then its logical negation is
$forall O in textrm{OS}: exists t: lnot c(O,t)$, by the usual rules for first order logic, which in English I would translate as "every operating system sometimes doesn't crash".
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
One way to look at the statement "some operating systems always crash":
$exists O in textrm{OS}: forall t: c(O,t)$, where I use OS for the set of operating systems and $t$ quantifies over times, $c(O,t)$ means the OS $O$ crashes at time $t$.
If you agree that this is the intended translation (this is a matter of linguistic discussion) then its logical negation is
$forall O in textrm{OS}: exists t: lnot c(O,t)$, by the usual rules for first order logic, which in English I would translate as "every operating system sometimes doesn't crash".
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
add a comment |
$begingroup$
One way to look at the statement "some operating systems always crash":
$exists O in textrm{OS}: forall t: c(O,t)$, where I use OS for the set of operating systems and $t$ quantifies over times, $c(O,t)$ means the OS $O$ crashes at time $t$.
If you agree that this is the intended translation (this is a matter of linguistic discussion) then its logical negation is
$forall O in textrm{OS}: exists t: lnot c(O,t)$, by the usual rules for first order logic, which in English I would translate as "every operating system sometimes doesn't crash".
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
$endgroup$
One way to look at the statement "some operating systems always crash":
$exists O in textrm{OS}: forall t: c(O,t)$, where I use OS for the set of operating systems and $t$ quantifies over times, $c(O,t)$ means the OS $O$ crashes at time $t$.
If you agree that this is the intended translation (this is a matter of linguistic discussion) then its logical negation is
$forall O in textrm{OS}: exists t: lnot c(O,t)$, by the usual rules for first order logic, which in English I would translate as "every operating system sometimes doesn't crash".
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
answered Feb 1 at 13:55
Henno BrandsmaHenno Brandsma
116k349127
116k349127
add a comment |
add a comment |
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$begingroup$
The negation of "some" is not "all". "Some" means "at least one". Thus, the negation of "some" is "not some", i.e. "all not".
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:33
$begingroup$
"always" is tricky here; if you have not a specific need to express the temporal fact, we can symply say : "Some operating systems always-crash", i.e. $exists x (text {OpSys}(x) land text {Crash}(x))$.
$endgroup$
– Mauro ALLEGRANZA
Feb 1 at 13:38