Negating a Symbolic Expression/Logic

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Negate the following and simplify as much as you can...
$$exists x ~forall y~(p(y) to forall z~q(z))$$
How would I negate this expression. Not sure how to start it.
discrete-mathematics logic
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up vote
4
down vote
favorite
Negate the following and simplify as much as you can...
$$exists x ~forall y~(p(y) to forall z~q(z))$$
How would I negate this expression. Not sure how to start it.
discrete-mathematics logic
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
Negate the following and simplify as much as you can...
$$exists x ~forall y~(p(y) to forall z~q(z))$$
How would I negate this expression. Not sure how to start it.
discrete-mathematics logic
Negate the following and simplify as much as you can...
$$exists x ~forall y~(p(y) to forall z~q(z))$$
How would I negate this expression. Not sure how to start it.
discrete-mathematics logic
discrete-mathematics logic
edited Sep 6 at 3:16
Graham Kemp
81.4k43275
81.4k43275
asked Sep 6 at 3:12
Vise
312
312
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1 Answer
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up vote
4
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Begin with placing a negation sign.
$qquadneg exists x ~forall y~(p(y) to forall z~q(z))$
Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.
$qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$
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1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
4
down vote
Begin with placing a negation sign.
$qquadneg exists x ~forall y~(p(y) to forall z~q(z))$
Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.
$qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$
add a comment |Â
up vote
4
down vote
Begin with placing a negation sign.
$qquadneg exists x ~forall y~(p(y) to forall z~q(z))$
Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.
$qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$
add a comment |Â
up vote
4
down vote
up vote
4
down vote
Begin with placing a negation sign.
$qquadneg exists x ~forall y~(p(y) to forall z~q(z))$
Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.
$qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$
Begin with placing a negation sign.
$qquadneg exists x ~forall y~(p(y) to forall z~q(z))$
Then use deMorgan's Rules for quantifier duality, and the negation of a conditional.
$qquadbeginalignneg exists x~phi ~&equiv~forall x~neg phi\[1ex]neg forall y~psi~&equiv~exists y~negpsi\[1ex]neg(chito xi)~&equiv~chiwedgenegxiendalign$
answered Sep 6 at 3:24
Graham Kemp
81.4k43275
81.4k43275
add a comment |Â
add a comment |Â
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