Are these logic statements equivalent?
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I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.
- $forall{x}.(P(x) iff Q(x))$
- $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$
- $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$
first-order-logic
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
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add a comment |
$begingroup$
I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.
- $forall{x}.(P(x) iff Q(x))$
- $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$
- $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$
first-order-logic
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
$endgroup$
add a comment |
$begingroup$
I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.
- $forall{x}.(P(x) iff Q(x))$
- $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$
- $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$
first-order-logic
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
$endgroup$
I understand logical statements #1 and #2 to be equivalent. I have been told that logical statement #3 is not equivalent to #2, but I do not understand how or why (assuming what I have been told is correct). The domain of variables x and y is the same.
- $forall{x}.(P(x) iff Q(x))$
- $forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x)))$
- $forall{x, y}.((P(x) to Q(x)) wedge (Q(y) to P(y)))$
first-order-logic
first-order-logic
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
asked Dec 2 '18 at 2:57
logically_confusedlogically_confused
31
31
add a comment |
add a comment |
1 Answer
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I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.
Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):
First, universal quantifiers distribute over conjunctions, so:
$$
forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
$$
Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
$$
(forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
$$
Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.
Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
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$begingroup$
I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.
Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):
First, universal quantifiers distribute over conjunctions, so:
$$
forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
$$
Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
$$
(forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
$$
Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.
Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
$endgroup$
add a comment |
$begingroup$
I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.
Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):
First, universal quantifiers distribute over conjunctions, so:
$$
forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
$$
Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
$$
(forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
$$
Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.
Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
$endgroup$
add a comment |
$begingroup$
I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.
Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):
First, universal quantifiers distribute over conjunctions, so:
$$
forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
$$
Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
$$
(forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
$$
Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.
Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
$endgroup$
I refuse to believe that #2 and #3 are not equivalent on the grounds that the explicit names of quantified variables don't matter, just that they are distinct from each other, e.g. for all univariate predicates $P$ we have that $forall x. P(x) iff forall y. P(y)$.
Here is an attempt to be as explicit as possible with a proof that they are equivalent (forgive me if it has a mistake, it has been a while since I've tried to write anything about predicate calculus):
First, universal quantifiers distribute over conjunctions, so:
$$
forall{x}.((P(x) to Q(x)) wedge (Q(x) to P(x))) iff (forall{x}.(P(x) to Q(x))) wedge (forall{x}.(Q(x) to P(x)))
$$
Second, conjunctive elimination will give us each half independently. Third, variable names don't matter so write $forall{x}.(Q(x) to P(x))$ as $forall{y}.(Q(y) to P(y))$. Fourth, conjunctive introduction gives us the following:
$$
(forall{x}.(P(x) to Q(x)))wedge (forall{y}.(Q(y) to P(y)))
$$
Finally, universal introduction of the other variable to each side, and factoring the quantifiers out of the conjunction, shows that #2 implies #3.
Now do something similar the other way around. Point being that this is obviously true, and only a sadist would make you prove it.
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
answered Dec 2 '18 at 3:48
FranklinBashFranklinBash
1012
1012
add a comment |
add a comment |
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