minimal vertex cover and P=NP

Question

could someone please explain to me why the following occurs?

let function f be a function that finds the minimal vertex cover. meaning: f(G,v)=minimal vertex cover that v belongs to

(the graph is indirected and no weights involved)

why if we know that f can be calculated in polynomial time then P=NP?

how do i prove such thing? i write because given f runs in polynomial time than P=NP?

i know that vertex cover is a NP-Hard(even NP-Complete). so how to write a proof? i just say since vertex cover is np complete, and minimal vertex cover is basically $min{f(x):x\in vertexcover(G)}$, then since i know it runs in polynomial time it means P=NP?

edit: from what i understand, the question is about reducing vertex cover to minimal vertex cover. since vertex cover is np-complete, then if it computes in polynomial time, then minimal vertex cover should also be computed in polynomial time. my problem is that reduction, i am not sure how to do it, how to do the reduction correctly.

would really appreciate to see how such a proof would look like to know what to aspire to in similar problems

thanks a lot

Answer 1

The class of NP-complete problems has a special property which has been proven :

If one problem can be solved in polynomial time, then all of them can be solved in polynomial time.

There must be a polynomial transformation from one problem to another in this class, but I am not aware of any details.

So, if you actually would find a polynomial algorithm for an NP-complete problem, you will become famous and rich.