First order logic.

Question

In Artificial intelligence, I saw the following question and answer in website.

Question:

Politicians can fool some people all of the time, and they can fool all people some of the time, but they can’t fool all of the people all of the time. $\newcommand{Politician}{\operatorname{Politician}}\newcommand{Person}{\operatorname{Person}}\newcommand{Fool}{\operatorname{Fool}}\newcommand{Time}{\operatorname{Time}}$

Answer:

Let $\Fool(x,y,t)$ mean that $x$ can fool $y$ at time $t$.
$$\begin{align} \forall x (\Politician(x) &\rightarrow\\ &(\exists y \Person(y) \rightarrow (\forall t \Time(t) \rightarrow \Fool(x,y,t))) \land\\ &(\exists t \Time(t) \rightarrow (\forall y \Person(y) \rightarrow \Fool(x,y,t)) \land \\ &(\exists t,y \Time(t) \land \Person(y) \rightarrow \lnot\Fool(x,y,t)) \end{align}$$

Could I write it in the following way. Is that correct?

My Solution:

$$ (\forall x\, \exists y\, \forall t\, (\Politician(x) \land \Person(y) \land \Fool(x,y,t)) \land\\ (\forall x\, \forall y\, \exists t\, (\Politician(x) \land \Person(y) \land \Fool(x,y,t)) \land\\ \lnot(\forall x\, \forall y\, \forall t\, (\Politician(x) \land \Person(y) \land \Fool(x,y,t))$$

Answer 1

Neither solution works correctly.

In your solution, for example, the conjunct $$\forall x: \exists y: \forall t: (\mathrm{Politician}(x) \land \mathrm{People}(y) \land \mathrm{fool}(x,y,t))$$ asserts that for any $x$ there exists an $y$ such that (among other things) $x$ is a politician. That is, you want everyone to be a politician -- and every time must be a politician too, because "$\forall x$" doesn't know that it is supposed to quantify only over people but not over times.

And your third conjunct $$\neg \forall x: \forall y: \forall t: (\mathrm{Politician}(x) \land \mathrm{People}(y) \land \mathrm{fool}(x,y,t))$$ will be true as long as there is anything that is not a politician or a person. Even if everyone can be fooled by anyone all the time, you can instantiate all of $x$, $y$ and $z$ to "Thursday". Then, because $\mathrm{People}(\mathrm{Thursday})$ is presumably false, the three nested quantifiers will all be false, and when you negate that you get True.

It should also count as suspicious that you're never even speaking about when something is a time or not.

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But the website answer you quote has its own problems. It seems that it tries to express "There exist a person with such-and-such property" as $$\exists y: \mathrm{person}(y) \Rightarrow \text{such-and-such}(y)$$ But because $\mathrm{false}\Rightarrow\mathit{anything}$ is always true, this condition is true as as soon as somewhere there exists anyone or anything that is not a person. Only if everything are persons (which shouldn't be the case here because times also exist) does the such-and-such property even influence the truth of that condition. The correct formulation would be $$\exists y: \mathrm{person}(y) \land \text{such-and-such}(y)$$

Answer 2

I agree with Henning, just wanted to point out, that natural language is not perfect, and there is more that one correct way of interpreting the phrase

Politicians can fool some people all of the time, and they can fool all people some of the time, but they can’t fool all of the people all of the time.

in first-order logic. The problem is that "fool some people all of the time" can be translated (as a part of a larger formula) to $$\forall t.\ \exists y.\ P(x,y,t)$$ or to $$\exists y.\ \forall t.\ Q(x,y,t)$$ and those two are not equivalent: first would mean that at every given time there is someone you can fool, and the second would mean that there is someone who at every given time can be fooled.

Answer 3

Comparing the first formula with the second, I see you moved some quantifiers (∀,∃) to the left, so I will assume this is what you are trying to do. As the others noted these 2 formulas are not equivalent. However it is certainly possible to seperate the predicate quantifiers from the rest of the proposition, at least in classical logic.

See prenex normal form.