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

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SUMMARY

The correct negation of the expression ( \exists x ) ( \forall y ) \Phi (x,y ) is \neg ( ( \exists x ) ( \forall y ) \Phi (x,y ) ) \equiv ( \forall x ) ( \exists y ) \neg \Phi (x,y ). This transformation follows logical equivalences, confirming that negating an existential quantifier leads to a universal quantifier and vice versa, while negating the predicate itself. The discussion also highlights the equivalence transformations involving nested quantifiers and negations, demonstrating the logical structure clearly.

PREREQUISITES
  • Understanding of first-order logic and quantifiers
  • Familiarity with logical equivalences and transformations
  • Knowledge of predicate logic notation
  • Basic skills in mathematical logic
NEXT STEPS
  • Study the principles of quantifier negation in mathematical logic
  • Learn about logical equivalences and their applications in proofs
  • Explore advanced topics in predicate logic and its applications
  • Practice transforming complex logical expressions and their negations
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Students of mathematics, logic enthusiasts, and anyone involved in formal reasoning or mathematical proofs will benefit from this discussion.

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

Is this correct?

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

Is this correct?

\neg ( ( \exists x ) ( \forall y ) \Phi (x,y ) ) \equiv ( \forall x ) ( \exists y ) \neg \Phi (x,y )

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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