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9 votes
9 votes

Let's make a trip to a new world called "Never Never Land".

Regular, ordinary first-order logic has two quantifiers: $\forall$ and $\exists$.

Now, let's imagine we lived in a world in which these quantifiers didn't exist, and instead we only had one quantifier, $\mathrm{N}$. The quantifier $\mathrm{N}$ is the "never" quantifier, and the expression
$$\mathrm{N} x.\; [\text{some formula}]$$
means "[some formula] is never true, regardless of what choice of $x$ we pick." For example, the expression
$\mathrm{Nx}(\mathrm{P}(\mathrm{x}))$ says "There is No element $\mathrm{x}$ in the domain, such that $\mathrm{P}(\mathrm{x})$ is true".

For predicates $\mathrm{A}(\mathrm{x})$ and $\mathrm{B}(\mathrm{x})$, Which of the following is the correct expression for $\text{“All A's are B's" }?$

  1. $\neg \mathrm{Nx}(\mathrm{A}(\mathrm{x}) \rightarrow \mathrm{B}(\mathrm{x}))$
  2. $\neg \mathrm{Nx}(\mathrm{A}(\mathrm{x}) \wedge \neg \mathrm{B}(\mathrm{x}))$
  3. $\mathrm{Nx}(\mathrm{A}(\mathrm{x}) \wedge \neg \mathrm{B}(\mathrm{x}))$
  4. $\mathrm{Nx}(\neg A(x) \wedge B(x))$
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2 Answers

6 votes
6 votes

This is a Stanford Question.

$\color{red}{\text{Detailed Video Solution:}}$ Detailed Video Solution with Complete Analysis

$$
\begin{aligned}
& \forall x \mathrm{P}(\mathrm{x})==\mathrm{Nx} \neg \mathrm{P}(\mathrm{x}) \\
& \exists \mathrm{xP}(\mathrm{x})== \neg \mathrm{NxP}(\mathrm{x})
\end{aligned}
$$
So, Option $C$ is the Correct answer.

edited by
Answer:

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