Negation of Propositional Function in Two Variables

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $\map P {x, y}$ be a propositional function of two Variables.

Then:

$\neg \forall x: \exists y: \map P {x, y} \iff \exists x: \forall y: \neg \map P {x, y}$


That is:

It is not the case that for all $x$ a value of $y$ can be found to satisfy $\map P {x, y}$

means the same thing as:

There exists at least one value of $x$ such that for all $y$ it is not possible to satisfy $\map P {x, y}$


Proof

\(\, \displaystyle \neg \forall x: \, \) \(\displaystyle \exists y\) \(:\) \(\displaystyle \map P {x, y}\)
\(\displaystyle \leadstoandfrom \ \ \) \(\, \displaystyle \exists x: \, \) \(\displaystyle \neg \exists y\) \(:\) \(\displaystyle \map P {x, y}\) Denial of Universality
\(\displaystyle \leadstoandfrom \ \ \) \(\, \displaystyle \exists x: \, \) \(\displaystyle \forall y\) \(:\) \(\displaystyle \neg \map P {x, y}\) Denial of Existence

$\blacksquare$


Sources