Mixing
Finite sequence of quantifiers “For all” and “there exists” in a mathematical statement, dependencies...
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The statement "$forallepsilonexistsdeltaforall xforall y$ $:P(epsilon,delta,x,y)$ is true" is equivalent to the same statement with $forall x$ and $forall y$ swapped.
And $delta$ can generally depend on $epsilon$.
If another statement is "$forall a_1exists a_2forall a_3exists a_4dotsexists a_{2n}:P(a_1,dots,a_{2n})$ is true"then how can we rearrange the quantifiers so that the statement is still true?
If there is a statement like $X_1 a_1 X_2 a_2...X_n a_n : P(a_1,dots,a_n)$ is true, where the $X_iin{forall,exists}$ and $a_i$ are some parameters, what permutations preserve the truthfulness? And which of the previously appearing operators does each operator generally depend on?
logic
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
$endgroup$
add a comment |
$begingroup$
The statement "$forallepsilonexistsdeltaforall xforall y$ $:P(epsilon,delta,x,y)$ is true" is equivalent to the same statement with $forall x$ and $forall y$ swapped.
And $delta$ can generally depend on $epsilon$.
If another statement is "$forall a_1exists a_2forall a_3exists a_4dotsexists a_{2n}:P(a_1,dots,a_{2n})$ is true"then how can we rearrange the quantifiers so that the statement is still true?
If there is a statement like $X_1 a_1 X_2 a_2...X_n a_n : P(a_1,dots,a_n)$ is true, where the $X_iin{forall,exists}$ and $a_i$ are some parameters, what permutations preserve the truthfulness? And which of the previously appearing operators does each operator generally depend on?
logic
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
$endgroup$
$begingroup$
Briefly --- Any permutation that doesn't swap the order of "there exists" and "for all" gives an equivalent statement, and any permutation that swaps the order of at least one pair of "there exists" and "for all" gives a possibly non-equivalent statement. The reason why I say "possibly non-equivalent" is because for certain special choices of $P$ you might get an equivalent statement, but there's always going to be some choice of $P$ for which the statements are not equivalent. Google prenex normal form.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:34
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The general rule is : you can swap the sime "type" of quantifier without changing the meaning of the formula (i.e. $forall x forall y$ is the same as $forall y forall x$, abd the same for $exists$) but you cannot exchange difefrent quantifiers (e.g. $forall x exists y$) without changing the meaning.
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:36
add a comment |
$begingroup$
The statement "$forallepsilonexistsdeltaforall xforall y$ $:P(epsilon,delta,x,y)$ is true" is equivalent to the same statement with $forall x$ and $forall y$ swapped.
And $delta$ can generally depend on $epsilon$.
If another statement is "$forall a_1exists a_2forall a_3exists a_4dotsexists a_{2n}:P(a_1,dots,a_{2n})$ is true"then how can we rearrange the quantifiers so that the statement is still true?
If there is a statement like $X_1 a_1 X_2 a_2...X_n a_n : P(a_1,dots,a_n)$ is true, where the $X_iin{forall,exists}$ and $a_i$ are some parameters, what permutations preserve the truthfulness? And which of the previously appearing operators does each operator generally depend on?
logic
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
$endgroup$
The statement "$forallepsilonexistsdeltaforall xforall y$ $:P(epsilon,delta,x,y)$ is true" is equivalent to the same statement with $forall x$ and $forall y$ swapped.
And $delta$ can generally depend on $epsilon$.
If another statement is "$forall a_1exists a_2forall a_3exists a_4dotsexists a_{2n}:P(a_1,dots,a_{2n})$ is true"then how can we rearrange the quantifiers so that the statement is still true?
If there is a statement like $X_1 a_1 X_2 a_2...X_n a_n : P(a_1,dots,a_n)$ is true, where the $X_iin{forall,exists}$ and $a_i$ are some parameters, what permutations preserve the truthfulness? And which of the previously appearing operators does each operator generally depend on?
logic
logic
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
asked Jan 22 at 13:57
John CataldoJohn Cataldo
1,1881316
1,1881316
$begingroup$
Briefly --- Any permutation that doesn't swap the order of "there exists" and "for all" gives an equivalent statement, and any permutation that swaps the order of at least one pair of "there exists" and "for all" gives a possibly non-equivalent statement. The reason why I say "possibly non-equivalent" is because for certain special choices of $P$ you might get an equivalent statement, but there's always going to be some choice of $P$ for which the statements are not equivalent. Google prenex normal form.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:34
$begingroup$
The general rule is : you can swap the sime "type" of quantifier without changing the meaning of the formula (i.e. $forall x forall y$ is the same as $forall y forall x$, abd the same for $exists$) but you cannot exchange difefrent quantifiers (e.g. $forall x exists y$) without changing the meaning.
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:36
add a comment |
$begingroup$
Briefly --- Any permutation that doesn't swap the order of "there exists" and "for all" gives an equivalent statement, and any permutation that swaps the order of at least one pair of "there exists" and "for all" gives a possibly non-equivalent statement. The reason why I say "possibly non-equivalent" is because for certain special choices of $P$ you might get an equivalent statement, but there's always going to be some choice of $P$ for which the statements are not equivalent. Google prenex normal form.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:34
$begingroup$
The general rule is : you can swap the sime "type" of quantifier without changing the meaning of the formula (i.e. $forall x forall y$ is the same as $forall y forall x$, abd the same for $exists$) but you cannot exchange difefrent quantifiers (e.g. $forall x exists y$) without changing the meaning.
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:36
$begingroup$
Briefly --- Any permutation that doesn't swap the order of "there exists" and "for all" gives an equivalent statement, and any permutation that swaps the order of at least one pair of "there exists" and "for all" gives a possibly non-equivalent statement. The reason why I say "possibly non-equivalent" is because for certain special choices of $P$ you might get an equivalent statement, but there's always going to be some choice of $P$ for which the statements are not equivalent. Google prenex normal form.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:34
$begingroup$
Briefly --- Any permutation that doesn't swap the order of "there exists" and "for all" gives an equivalent statement, and any permutation that swaps the order of at least one pair of "there exists" and "for all" gives a possibly non-equivalent statement. The reason why I say "possibly non-equivalent" is because for certain special choices of $P$ you might get an equivalent statement, but there's always going to be some choice of $P$ for which the statements are not equivalent. Google prenex normal form.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:34
$begingroup$
The general rule is : you can swap the sime "type" of quantifier without changing the meaning of the formula (i.e. $forall x forall y$ is the same as $forall y forall x$, abd the same for $exists$) but you cannot exchange difefrent quantifiers (e.g. $forall x exists y$) without changing the meaning.
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:36
$begingroup$
The general rule is : you can swap the sime "type" of quantifier without changing the meaning of the formula (i.e. $forall x forall y$ is the same as $forall y forall x$, abd the same for $exists$) but you cannot exchange difefrent quantifiers (e.g. $forall x exists y$) without changing the meaning.
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:36
add a comment |
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$begingroup$
Briefly --- Any permutation that doesn't swap the order of "there exists" and "for all" gives an equivalent statement, and any permutation that swaps the order of at least one pair of "there exists" and "for all" gives a possibly non-equivalent statement. The reason why I say "possibly non-equivalent" is because for certain special choices of $P$ you might get an equivalent statement, but there's always going to be some choice of $P$ for which the statements are not equivalent. Google prenex normal form.
$endgroup$
– Dave L. Renfro
Jan 22 at 14:34
$begingroup$
The general rule is : you can swap the sime "type" of quantifier without changing the meaning of the formula (i.e. $forall x forall y$ is the same as $forall y forall x$, abd the same for $exists$) but you cannot exchange difefrent quantifiers (e.g. $forall x exists y$) without changing the meaning.
$endgroup$
– Mauro ALLEGRANZA
Jan 22 at 14:36