If some person is tall, then every person is tall.
Is this written as
$$\exists x(P(x) \land T(x)) \to \forall x(P(x) \to T(x))$$
or
$$\exists x(P(x) \land T(x)) \to \forall y(P(y) \to T(y))$$
or something completely different?
If some person is tall, then every person is tall.
Is this written as
$$\exists x(P(x) \land T(x)) \to \forall x(P(x) \to T(x))$$
or
$$\exists x(P(x) \land T(x)) \to \forall y(P(y) \to T(y))$$
or something completely different?
On
I would do it this way:
Denote the set of people by $X$ and the statement "$x$ is tall" by $T(x)$. Then:
$$\exists x\in X : T(x)\implies T(y)\:\forall y\in X$$
On
If some person is tall, then every person is tall.
Do you mean, "If there is some person who is tall, then every person is tall"?
$$\tag{1} \Big(\exists x \big(P(x)\wedge T(x)\big)\Big) \to \Big(\forall y \big(P(y)\to T(y)\big)\Big)\\ \equiv \\ \Big(\exists x \in P:T(x)\Big) \to \Big(\forall y\in P: T(y)\Big)$$
Or do you mean "There is some person who, if they are tall then every person is tall." (vis the 'Drinker's Paradox')
$$\tag{2} \exists x \Big(P(x)\wedge \big(T(x) \to \forall y \big(P(y)\to T(y)\big)\big)\Big) \\ \equiv \\ \exists x\in P:\Big(T(x) \to \big(\forall y\in P:T(y)\big)\Big) $$
The scope brackets are important here. (1) and (2) are quite different statements.
"There exists a person who, if he is is tall, then every person is tall."
(*) $\exists x \in P \ (T(x) \implies \forall x \in P, \ T(x))$
Interestingly, this statement is true. There is some person in the world such that if that person is tall, then everyone is tall. How can this seemingly absurd statement be true?
Well, suppose that there exists some person $x$ who is not tall. Then $T(x)$ is false, making the implication $T(x) \implies \forall x \in P, \ T(x))$ true.
But suppose every person is tall. Then the consequent of (*), $\forall x \in P, \ T(x))$, is true; making the implication true.
This story is known as the Drinker's paradox. In every (nonempty) bar, there's someone such that if he is drinking, then everyone is drinking. (And of course an empty bar would go out of business so we need not consider that case :-))