Writing “Someone has visited every country in the world except Libya” using quantifiers.
$begingroup$
Someone has visited every country in the world except Libya.
Let $B(x,y)-x,has,visited,country,y$
Domain for x is all people and for y is all countries
is $exists x forall y [B(x,y) land (y neq Libya)]$
the correct representation?
predicate-logic
edited Dec 9 '18 at 4:43
Shaun
8,941113681
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
$endgroup$
add a comment |
$begingroup$
Someone has visited every country in the world except Libya.
Let $B(x,y)-x,has,visited,country,y$
Domain for x is all people and for y is all countries
is $exists x forall y [B(x,y) land (y neq Libya)]$
the correct representation?
predicate-logic
edited Dec 9 '18 at 4:43
Shaun
8,941113681
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
$endgroup$
add a comment |
$begingroup$
Someone has visited every country in the world except Libya.
Let $B(x,y)-x,has,visited,country,y$
Domain for x is all people and for y is all countries
is $exists x forall y [B(x,y) land (y neq Libya)]$
the correct representation?
predicate-logic
edited Dec 9 '18 at 4:43
Shaun
8,941113681
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
$endgroup$
Someone has visited every country in the world except Libya.
Let $B(x,y)-x,has,visited,country,y$
Domain for x is all people and for y is all countries
is $exists x forall y [B(x,y) land (y neq Libya)]$
the correct representation?
predicate-logic
predicate-logic
edited Dec 9 '18 at 4:43
Shaun
8,941113681
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
edited Dec 9 '18 at 4:43
Shaun
8,941113681
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
edited Dec 9 '18 at 4:43
Shaun
8,941113681
edited Dec 9 '18 at 4:43
Shaun
8,941113681
edited Dec 9 '18 at 4:43
Shaun
8,941113681
8,941113681
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
asked Dec 9 '18 at 4:32
user3767495user3767495
3938
3938
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Your original statement implies that the someone in question has not visited Libya. That is, we need $B(x$, Libya$)$ to be false, but your answer doesn't imply that. According to your answer, $B(x$, Libya$)$ may be true or false, since the expression will evaluate to false because of $yne$ Libya regardless. You can modify it as:
$exists x[forall y (B(x,y) land (y neq $Libya$))]wedge neg B(x$, Libya$)$
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
Very subtle observation.
$endgroup$
– alex811
Dec 9 '18 at 5:21
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
$begingroup$
Your original statement implies that the someone in question has not visited Libya. That is, we need $B(x$, Libya$)$ to be false, but your answer doesn't imply that. According to your answer, $B(x$, Libya$)$ may be true or false, since the expression will evaluate to false because of $yne$ Libya regardless. You can modify it as:
$exists x[forall y (B(x,y) land (y neq $Libya$))]wedge neg B(x$, Libya$)$
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
Very subtle observation.
$endgroup$
– alex811
Dec 9 '18 at 5:21
add a comment |
$begingroup$
Your original statement implies that the someone in question has not visited Libya. That is, we need $B(x$, Libya$)$ to be false, but your answer doesn't imply that. According to your answer, $B(x$, Libya$)$ may be true or false, since the expression will evaluate to false because of $yne$ Libya regardless. You can modify it as:
$exists x[forall y (B(x,y) land (y neq $Libya$))]wedge neg B(x$, Libya$)$
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
$endgroup$
$begingroup$
Very subtle observation.
$endgroup$
– alex811
Dec 9 '18 at 5:21
add a comment |
$begingroup$
Your original statement implies that the someone in question has not visited Libya. That is, we need $B(x$, Libya$)$ to be false, but your answer doesn't imply that. According to your answer, $B(x$, Libya$)$ may be true or false, since the expression will evaluate to false because of $yne$ Libya regardless. You can modify it as:
$exists x[forall y (B(x,y) land (y neq $Libya$))]wedge neg B(x$, Libya$)$
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
$endgroup$
Your original statement implies that the someone in question has not visited Libya. That is, we need $B(x$, Libya$)$ to be false, but your answer doesn't imply that. According to your answer, $B(x$, Libya$)$ may be true or false, since the expression will evaluate to false because of $yne$ Libya regardless. You can modify it as:
$exists x[forall y (B(x,y) land (y neq $Libya$))]wedge neg B(x$, Libya$)$
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
edited Dec 9 '18 at 4:57
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
answered Dec 9 '18 at 4:52
Shubham JohriShubham Johri
4,992717
4,992717
$begingroup$
Very subtle observation.
$endgroup$
– alex811
Dec 9 '18 at 5:21
add a comment |
$begingroup$
Very subtle observation.
$endgroup$
– alex811
Dec 9 '18 at 5:21
$begingroup$
Very subtle observation.
$endgroup$
– alex811
Dec 9 '18 at 5:21
$begingroup$
Very subtle observation.
$endgroup$
– alex811
Dec 9 '18 at 5:21
add a comment |
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