Equivalence of tuple relational calculus expression
A. $forall t in r left(Pleft(tright)right)$
B. $exists t notin r left(neg Pleft(tright)right)$
solving B,
$negforall tin r left(neg Pleft(tright)right)$
$negforall tnotin r left(Pleft(tright)right)$
$neg neg forall t in r left(Pleft(tright)right)$
$forall t in r left(Pleft(tright)right)$
which tells us A and B are equivalent, but I'm not sure that negation propagates through $in$ like the way I did above. Please explain how $neg$ propagates through $in$.
first-order-logic quantifiers
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
add a comment |
A. $forall t in r left(Pleft(tright)right)$
B. $exists t notin r left(neg Pleft(tright)right)$
solving B,
$negforall tin r left(neg Pleft(tright)right)$
$negforall tnotin r left(Pleft(tright)right)$
$neg neg forall t in r left(Pleft(tright)right)$
$forall t in r left(Pleft(tright)right)$
which tells us A and B are equivalent, but I'm not sure that negation propagates through $in$ like the way I did above. Please explain how $neg$ propagates through $in$.
first-order-logic quantifiers
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
add a comment |
A. $forall t in r left(Pleft(tright)right)$
B. $exists t notin r left(neg Pleft(tright)right)$
solving B,
$negforall tin r left(neg Pleft(tright)right)$
$negforall tnotin r left(Pleft(tright)right)$
$neg neg forall t in r left(Pleft(tright)right)$
$forall t in r left(Pleft(tright)right)$
which tells us A and B are equivalent, but I'm not sure that negation propagates through $in$ like the way I did above. Please explain how $neg$ propagates through $in$.
first-order-logic quantifiers
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
A. $forall t in r left(Pleft(tright)right)$
B. $exists t notin r left(neg Pleft(tright)right)$
solving B,
$negforall tin r left(neg Pleft(tright)right)$
$negforall tnotin r left(Pleft(tright)right)$
$neg neg forall t in r left(Pleft(tright)right)$
$forall t in r left(Pleft(tright)right)$
which tells us A and B are equivalent, but I'm not sure that negation propagates through $in$ like the way I did above. Please explain how $neg$ propagates through $in$.
first-order-logic quantifiers
first-order-logic quantifiers
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
edited Dec 1 '18 at 11:03
Martin Sleziak
44.7k8115271
44.7k8115271
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
asked Dec 1 '18 at 5:07
Mk UtkarshMk Utkarsh
89110
89110
add a comment |
add a comment |
1 Answer
1
active
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votes
First of all the syntax is incorrect: either choose lower-case for sets and upper-case for element, or vise-verse. It is not grammatical to say $xin t$ or $tin r$. That is nit-picky, but it helps in rigor.
That said, Your very first step after solving B is wrong, refer to the what was shown above.
Semantic Equivalence:
To say $forall t in r(P(t))$ is to say every $t$ in $r$ is $P$ or generally symbolized $P(t)$
Similarly, to say $exists t notin r(lnot P(t))$ is to say there is a $t$ not in $r$ such that it is not $P$ or $P(t)$.
The two are not equivalent: One is saying every $t$ in $r$ is $P$, the other is saying there is a $t$ not in $r$ which is not $P$.
Syntactic Equivalence between $forall x(xin Y)$ and $exists x(xin Y)$:
Let $Q(x):xin Y$; therefore, $forall x(Q(x))equiv lnot exists x(lnot Q(x))$, but $lnot Q(x)equivlnot(xin Y)equiv xnotin Y$; therefore, $forall x(Q(x))equivlnot exists x(xnotin Y)$.
Let $Q(x):xin Y$; therefore, $exists x Q(x)equiv lnot forall x(lnot Q(x))$, but $lnot Q(x)equiv lnot (xin Y)equiv xnotin Y$; therefore, $exists x(xin Y)equivlnot forall x(xnotin Y)$
These are the only ways negation can propogate through $in$
Keeping that in mind: $forall t in r(P(t))equiv lnot exists tin r(lnot P(t))$ and not $forall t in r(P(t))not equiv exists tnotin r(lnot P(t))$
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
add a comment |
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1 Answer
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1 Answer
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active
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votes
First of all the syntax is incorrect: either choose lower-case for sets and upper-case for element, or vise-verse. It is not grammatical to say $xin t$ or $tin r$. That is nit-picky, but it helps in rigor.
That said, Your very first step after solving B is wrong, refer to the what was shown above.
Semantic Equivalence:
To say $forall t in r(P(t))$ is to say every $t$ in $r$ is $P$ or generally symbolized $P(t)$
Similarly, to say $exists t notin r(lnot P(t))$ is to say there is a $t$ not in $r$ such that it is not $P$ or $P(t)$.
The two are not equivalent: One is saying every $t$ in $r$ is $P$, the other is saying there is a $t$ not in $r$ which is not $P$.
Syntactic Equivalence between $forall x(xin Y)$ and $exists x(xin Y)$:
Let $Q(x):xin Y$; therefore, $forall x(Q(x))equiv lnot exists x(lnot Q(x))$, but $lnot Q(x)equivlnot(xin Y)equiv xnotin Y$; therefore, $forall x(Q(x))equivlnot exists x(xnotin Y)$.
Let $Q(x):xin Y$; therefore, $exists x Q(x)equiv lnot forall x(lnot Q(x))$, but $lnot Q(x)equiv lnot (xin Y)equiv xnotin Y$; therefore, $exists x(xin Y)equivlnot forall x(xnotin Y)$
These are the only ways negation can propogate through $in$
Keeping that in mind: $forall t in r(P(t))equiv lnot exists tin r(lnot P(t))$ and not $forall t in r(P(t))not equiv exists tnotin r(lnot P(t))$
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
add a comment |
First of all the syntax is incorrect: either choose lower-case for sets and upper-case for element, or vise-verse. It is not grammatical to say $xin t$ or $tin r$. That is nit-picky, but it helps in rigor.
That said, Your very first step after solving B is wrong, refer to the what was shown above.
Semantic Equivalence:
To say $forall t in r(P(t))$ is to say every $t$ in $r$ is $P$ or generally symbolized $P(t)$
Similarly, to say $exists t notin r(lnot P(t))$ is to say there is a $t$ not in $r$ such that it is not $P$ or $P(t)$.
The two are not equivalent: One is saying every $t$ in $r$ is $P$, the other is saying there is a $t$ not in $r$ which is not $P$.
Syntactic Equivalence between $forall x(xin Y)$ and $exists x(xin Y)$:
Let $Q(x):xin Y$; therefore, $forall x(Q(x))equiv lnot exists x(lnot Q(x))$, but $lnot Q(x)equivlnot(xin Y)equiv xnotin Y$; therefore, $forall x(Q(x))equivlnot exists x(xnotin Y)$.
Let $Q(x):xin Y$; therefore, $exists x Q(x)equiv lnot forall x(lnot Q(x))$, but $lnot Q(x)equiv lnot (xin Y)equiv xnotin Y$; therefore, $exists x(xin Y)equivlnot forall x(xnotin Y)$
These are the only ways negation can propogate through $in$
Keeping that in mind: $forall t in r(P(t))equiv lnot exists tin r(lnot P(t))$ and not $forall t in r(P(t))not equiv exists tnotin r(lnot P(t))$
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
add a comment |
First of all the syntax is incorrect: either choose lower-case for sets and upper-case for element, or vise-verse. It is not grammatical to say $xin t$ or $tin r$. That is nit-picky, but it helps in rigor.
That said, Your very first step after solving B is wrong, refer to the what was shown above.
Semantic Equivalence:
To say $forall t in r(P(t))$ is to say every $t$ in $r$ is $P$ or generally symbolized $P(t)$
Similarly, to say $exists t notin r(lnot P(t))$ is to say there is a $t$ not in $r$ such that it is not $P$ or $P(t)$.
The two are not equivalent: One is saying every $t$ in $r$ is $P$, the other is saying there is a $t$ not in $r$ which is not $P$.
Syntactic Equivalence between $forall x(xin Y)$ and $exists x(xin Y)$:
Let $Q(x):xin Y$; therefore, $forall x(Q(x))equiv lnot exists x(lnot Q(x))$, but $lnot Q(x)equivlnot(xin Y)equiv xnotin Y$; therefore, $forall x(Q(x))equivlnot exists x(xnotin Y)$.
Let $Q(x):xin Y$; therefore, $exists x Q(x)equiv lnot forall x(lnot Q(x))$, but $lnot Q(x)equiv lnot (xin Y)equiv xnotin Y$; therefore, $exists x(xin Y)equivlnot forall x(xnotin Y)$
These are the only ways negation can propogate through $in$
Keeping that in mind: $forall t in r(P(t))equiv lnot exists tin r(lnot P(t))$ and not $forall t in r(P(t))not equiv exists tnotin r(lnot P(t))$
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
First of all the syntax is incorrect: either choose lower-case for sets and upper-case for element, or vise-verse. It is not grammatical to say $xin t$ or $tin r$. That is nit-picky, but it helps in rigor.
That said, Your very first step after solving B is wrong, refer to the what was shown above.
Semantic Equivalence:
To say $forall t in r(P(t))$ is to say every $t$ in $r$ is $P$ or generally symbolized $P(t)$
Similarly, to say $exists t notin r(lnot P(t))$ is to say there is a $t$ not in $r$ such that it is not $P$ or $P(t)$.
The two are not equivalent: One is saying every $t$ in $r$ is $P$, the other is saying there is a $t$ not in $r$ which is not $P$.
Syntactic Equivalence between $forall x(xin Y)$ and $exists x(xin Y)$:
Let $Q(x):xin Y$; therefore, $forall x(Q(x))equiv lnot exists x(lnot Q(x))$, but $lnot Q(x)equivlnot(xin Y)equiv xnotin Y$; therefore, $forall x(Q(x))equivlnot exists x(xnotin Y)$.
Let $Q(x):xin Y$; therefore, $exists x Q(x)equiv lnot forall x(lnot Q(x))$, but $lnot Q(x)equiv lnot (xin Y)equiv xnotin Y$; therefore, $exists x(xin Y)equivlnot forall x(xnotin Y)$
These are the only ways negation can propogate through $in$
Keeping that in mind: $forall t in r(P(t))equiv lnot exists tin r(lnot P(t))$ and not $forall t in r(P(t))not equiv exists tnotin r(lnot P(t))$
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
edited Dec 1 '18 at 9:25
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
answered Dec 1 '18 at 7:18
Bertrand Wittgenstein's GhostBertrand Wittgenstein's Ghost
354114
354114
add a comment |
add a comment |
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