B Translating Abraham Lincoln quote into symbolic logic

[Eclair_de_XII]

TL;DR Summary

Express the following in symbolic logic, and then write their respective negations.

1. You can fool all of the people some of the time.
2. You can fool some of the people all of the time.
3. But you cannot fool all of the people all of the time.

Can anybody check my work regarding these three statements, the third, in particular, please? I think I got the first two statements down, but I think that I'd feel safer if I got a second opinion. I think I also have a correct translation of the third statement down, but only because I understood that its negation translates to what you'd (read:fig) expect in common English. I originally had another answer for the third statement, but realized that its negation was not exactly correct. I am unsure about how to translate my original answer into English.

\begin{eqnarray}
m:=\textrm{moment}\\
T:=\textrm{time}\\
P:=\textrm{set of people}\\
f(x):=\textrm{You can fool }x
\end{eqnarray}

===1===
$\exists m\in T,\forall x\in P, f(x)$
There are moments in time s.t. you can fool every person (in the set of people).

$\forall m\in T,\exists x\in P,\lnot f(x)$
For every moment in time, there is a person whom you cannot fool.

===2===
$\exists x\in P, \forall m\in T, f(x)$
There are people whom you can fool for all moments in time.

$\forall x\in P,\exists m\in T,\lnot f(x)$
All people have a moment in time when they cannot be fooled.

===3===
$\forall m\in T,\lnot (\exists x\in P, \lnot f(x))$
For every moment in time, there is not a person whom you cannot fool.
Original answer: $\forall m\in T,\lnot\exists x\in P, \lnot f(x)$

$\exists m\in T,\exists x\in P,\lnot f(x)$
There is a moment in time during which there is a person that cannot be fooled.

 

[mjc123]

3 is the negation of what Lincoln said.

 

[PeroK]

I know nothing about this stuff, but surely you need something like $f(x, T):=$ you can fool person $x$ at time $T$?

 

[nuuskur]

Second that. The fooling-people-operator depends on two quantities.

 

[Eclair_de_XII]

This is helpful. I think that nixing every instance of $m\in$, and replacing every instance of $f(x)$ with $f(x,T)$, would suffice to correct for this?

 

[Office_Shredder]

I don't understand why you wrote two statements for part 1 and 2. Don't you just need the first one each time?

 

[PeroK]

I don't understand why you wrote two statements for part 1 and 2. Don't you just need the first one each time?

He had to give the negation as well.

 

[Office_Shredder]

Whoops, missed that part.

 

[sysprog]

From https://faculty.math.illinois.edu/~rdeville/teaching/347/hw2S.pdf 4. 2.19.

Let P be the set of all people, let T be the set of all times, and let F(p, t) be true if it is possible to fool person p at time t. The first clause of the sentence is “You can fool all of the people some of the time”, which translated into logic becomes $$(∃t ∈ T )(∀p ∈ P)F(p, t).$$ Here is one place to be careful! What Lincoln is saying is that there are times that you can fool everyone, which is what we have written mathematically. The statement $$(∀p ∈ P)(∃t ∈ T )F(p, t)$$ is not what he meant, for this is saying that for any given person, there is some time at which they can be fooled. The second clause of the sentence is “You can fool some of the people all of the time”. The way to translate this in logic is to say $$(∃p ∈ P)(∀t ∈ T )F(p, t),$$ meaning that there are people out there who can always be fooled. Again, flipping the quantifiers gives the wrong sense, because the statement $$(∀t ∈ T )(∃p ∈ P)F(p, t)$$ means that at any given time, there are some people to be fooled, but these could be different people every time you try it — and he’s saying there are some people who can always be fooled. Finally, the last clause is “You can’t fool all of the people all of the time”, which is then $$¬(∀p ∈ P)(∀t ∈ T )F(p, t).$$ This case is simplest since it doesn’t matter which order quantifiers come in if they are the same type of quantifier. Also, we can push the ¬ past the quantifiers, and a logically equivalent statement is $$(∃p ∈ P)(∃t ∈ T )¬F(p, t).$$ This means, of course, that there is at least one given person and one given time such that that person cannot be fooled then. So, putting Lincoln’s whole statement together gives $$((∃t ∈ T )(∀p ∈ P)F(p, t)) ∧ ((∃p ∈ P)(∀t ∈ T )F(p, t)) ∧ ((∃p ∈ P)(∃t ∈ T )¬F(p, t))$$ To do the negation, we need two rules: the first is the one we already used to push a ¬ past a quantifier, and the second is the DeMorgan’s Law which says $$¬(P ∧ Q ∧ R) = (¬P) ∨ (¬Q) ∨ (¬R).$$ So we have $$¬((∃t ∈ T )(∀p ∈ P)F(p, t)) ∧$$
$$((∃p ∈ P)(∀t ∈ T )F(p, t)) ∧$$
$$((∃p ∈ P)(∃t ∈ T )¬F(p, t)) (¬(∃t ∈ T )(∀p ∈ P)F(p, t)) ∨$$
$$(¬(∃p ∈ P)(∀t ∈ T )F(p, t)) ∨$$
$$(¬(∃p ∈ P)(∃t ∈ T )¬F(p, t)) ((∀t ∈ T )(∃p ∈ P)¬F(p, t)) ∨$$
$$((∀p ∈ P)(∃t ∈ T )¬F(p, t)) ∨ ((∀p ∈ P)(∀t ∈ T )F(p, t)).$$ The English translation of this is “One of the following three things is true: either it’s true that at any given time, there are people who cannot be fooled, or it’s true that every person has times at which they cannot be true, or it’s true that everyone can be fooled at all times.” My feeling is that Lincoln is right, and this negation is wrong.