p(x)⇒∀x.p(x) is contingent?

Question

I've encountered a question asking whether the flowing sentence is valid/contingent/unsatisfiable:

p(x)⇒∀x.p(x) 

I think the answer is the sentence is valid. under section 6.10 of the textbook here http://logic.stanford.edu/intrologic/secondary/notes/chapter_06.htmlsays

a sentence with free variables is equivalent to the sentence in which all of the free variables are universally quantified.

Therefore I think the first relational sentence p(x) is equal to ∀x.p(x) and therefore the sentence is valid, ie. it is always true.

However,the correct answer is that the sentence is contingent viz. under some truth assignment it is true and other some other truth assignment it is false.

So why is the sentence contingent?Is the answer wrong?

Answer 1

You have a statement:

p(x)⇒∀x.p(x)

If you universally close the free variable, you get:

∀x.(p(x)⇒∀x.p(x))

in other words:

∀x.(p(x)⇒∀y.p(y))

which is not tautology, but is contingent. In non-technical terms, this reads:

for any x, if p(x) is true, then p(y) is true for all y

or, to transform it into an equivalent form:

(∃x.p(x))⇒(∀y.p(y))

it reads:

if p(x) is true for some x, then p(y) is true for all y

In other words,

p(x) is either always true or always false