Can anyone help me understand why this might be wrong?
You can’t fool all of the people all of the time.
(∀x) [P(x) /\ (∀y)(T(y)-> ~ F(x,y))]
Can anyone help me understand why this might be wrong?
You can’t fool all of the people all of the time.
(∀x) [P(x) /\ (∀y)(T(y)-> ~ F(x,y))]
On
A counter-example to "you CAN fool ALL of the people ALL of the time" would need to be of the form: there exists both a person x, and a time y, such that person x is not fooled at time y:
$$\exists x (\exists y ((P(x) \wedge T(y) \wedge \neg F(x,y)))$$
Of course, as Lincoln also asserted (which need not contradict the above):
$$\exists x (\forall y ((P(x) \wedge T(y)) \rightarrow F(x,y)))$$
"there exists a person who can be fooled for all times". And
$$\exists y (\forall x ((P(x) \wedge T(y) \rightarrow F(x,y)))$$
"there exists a time for which all persons are fooled". But as in your question, he finally asserts:
$$\neg(\forall x (\forall y ((P(x) \wedge (T(y)) \rightarrow F(x,y)))$$
which is equivalent to the asserting that the above counter-example is true.
You have: $\forall x \Bigg(P(x) \wedge \forall y\; \Big(T(y) \to \neg F(x,y)\Big)\Bigg)$
Which says: "All $x$ are people and for any $y$ that is a time then you can't fool that person at that time."
Basically: "Everything is a person that cannot be fooled at any time."
You want: $\forall x \Bigg(P(x) \to \neg \forall y\;\Big(T(y) \to F(x,y)\Big)\Bigg)$
"For any $x$ if it is a person then not any $y$ if that is a time then you can fool that person at that time"
Or equivalently: $\forall x \Bigg(P(x) \to \exists y\;\Big(T(y) \wedge \neg F(x,y)\Big)\Bigg)$
"For any $x$ that is a person then there is a $y$ that is a time and you cannot fool that person at that time."
Or even: $\exists x \Bigg(P(x)\wedge \exists y \Big(T(y) \wedge \neg F(x,y)\Big)\Bigg)$
"There is $x$ that is a person and $y$ that is a time and you cannot fool that person at that time."
Which all mean: "You cannot fool all the people all the time"
$$\neg \Bigg(\forall x \Big(P(x) \to \forall y \big(T(y) \to F(x,y)\big)\Big)\Bigg)$$
Keynotes:
To state "for every thing in a category there is a happening" use $\forall x (C(x)\to H(x))$
To state "for some thing in a category there is a happening" use $\exists x (C(x)\wedge H(x))$
Then the statement "Not for every thing in a category there is a happening" becomes "for some thing in a category there is a not-happening"
$$\neg\forall x(C(x)\to H(x)) \equiv \exists x (C(x) \wedge \neg H(x))$$