Negating a Math Equation: ( \exists x ) ( \forall y ) \Phi (x,y )

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Discussion Overview

The discussion centers on the negation of the mathematical expression ( \exists x ) ( \forall y ) \Phi (x,y ). Participants are exploring the correct formulation of its negation, which involves logical quantifiers and their transformations.

Discussion Character

  • Mathematical reasoning

Main Points Raised

  • One participant proposes that the negation is given by \neg ( ( \exists x ) ( \forall y ) \Phi (x,y ) ) \equiv ( \forall x ) ( \exists y ) \neg \Phi (x,y ).
  • Another participant agrees with the initial proposal, indicating that it looks correct.
  • A subsequent post reiterates the negation and provides a series of equivalences involving logical transformations, suggesting that the negation can be expressed as (Ax)(Ey)~F(x,y).

Areas of Agreement / Disagreement

Participants generally agree on the correctness of the negation presented, with no significant disagreement noted in the posts.

Contextual Notes

The discussion does not address any limitations or unresolved mathematical steps explicitly.

Who May Find This Useful

Readers interested in mathematical logic, particularly in the manipulation of quantifiers and negation in formal expressions.

gnome
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I want to negate this: [itex]( \exists x ) ( \forall y ) \Phi (x,y )[/itex]

Is this correct?

[tex]\neg ( ( \exists x ) ( \forall y ) \Phi (x,y ) ) \equiv ( \forall x ) ( \exists y ) \neg \Phi (x,y )[/tex]
 
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looks right
 
gnome said:
I want to negate this: [itex]( \exists x ) ( \forall y ) \Phi (x,y )[/itex]

Is this correct?

[tex]\neg ( ( \exists x ) ( \forall y ) \Phi (x,y ) ) \equiv ( \forall x ) ( \exists y ) \neg \Phi (x,y )[/tex]

Yes.

~[(Ex)(Ay)F(x,y)] <-> ~(Ex)(Ay)F(x.y)
~(Ex)(Ay)F(x,y) <-> (Ax)~(Ay)F(x,y)
(Ax)~(Ay)F(x,y) <-> (Ax)(Ey)~F(x,y)
therefore,
~[(Ex)(Ay)F(x,y)] <-> (Ax)(Ey)~F(x,y).
 
Thanks guys.
 

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