Discovering the Useful Denial for Predicates P(x) and Q(x) in Universe U

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SUMMARY

The discussion focuses on deriving a useful denial for the logical expression (∀x∈U)(P(x)⇒Q(x)). The correct denial is established as ¬(∀x∈U)(P(x)⇒Q(x)), which simplifies to (∃x∈U)¬(P(x)⇒Q(x)). The final clarification provided indicates that ¬(P(x)⇒Q(x)) is equivalent to the existence of an element x0 in U such that P(x0) is true and Q(x0) is false, represented as P(x0) ∧ ¬Q(x0).

PREREQUISITES
  • Understanding of predicate logic and quantifiers
  • Familiarity with logical implications and their negations
  • Knowledge of logical equivalences in propositional logic
  • Basic concepts of set theory and universes in mathematical logic
NEXT STEPS
  • Study logical equivalences in propositional logic
  • Learn about the application of quantifiers in predicate logic
  • Explore the concept of useful denials in logical expressions
  • Investigate examples of predicates and their implications in mathematical proofs
USEFUL FOR

Students of mathematics, logicians, and anyone interested in the foundations of logical reasoning and predicate calculus.

ver_mathstats
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Let U be universe under consideration, and let P(x) and Q(x) be predicate with free variable x. Find a useful denial for (∀x∈U)(P(x)⇒Q(x))

Then my answer is:

¬(∀x∈U)(P(x)⇒Q(x))

(∃x∈U)¬(P(x)⇒Q(x))

¬(P(x)⇒Q(x))

¬P∨Q

I'm unsure if I am on the write path when it comes to finding the useful denial, is this how to do it?

Thank you.
 
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Your last step is wrong.
$$
\lnot \,(P(x) \Longrightarrow Q(x)) \Longleftrightarrow \exists \,x_0 \in U \, : \,P(x_0) \wedge \lnot \,Q(x_0)
$$
 

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