(rev. May 12, 2022)
VC ≤P IS
VC is the problem: Does a graph have a vertex cover of size d or less?
IS is the problem: Does a graph have an independent set of size k or more?
Claim: VC ≤P IS
Proof:
( We will use the fact that the complement of a vertex cover of a graph is an independent set of nodes, and conversely the complement of an independent set of nodes is a vertex cover.)
Let x be a problem that is an instance of VC. In other words x = the question: Is there a vertex cover of the graph G = (V,E) of size ≤ d (where d ≥ 1 is some integer)?
Let AIS be an algorithm that solves IS. Submit this problem to AIS: y = the question: Is there an independent set of nodes in the graph G of size ≥ n-d (where n=|V|)? (Here we are doing a polynomial amount of work to construct y from x.)
If AIS outputs "yes" for the input y, then ∃ an independent subset S ⊂ V of V of size ≥ n-d.
Therefore the complement of S, V-S is a vertex cover of G of size ≤ d.
Therefore the answer to x is also "yes." (We can do a small amount of work, and translate yes answer from AIS into a yes answer for problem x.)
If AIS outputs "no" for the input y, then ∄ an independent subset S ⊂ V of V of size ≥ n-d.
Therefore ∄ a complement of such an independent subset S ⊂ V of V of size ≥ n-d. Therefore there is no vertex cover of G of size ≤ d.
Therefore the answer to x is also "no." (We can do a small amount of work, and translate no answer from AIS into a no answer for problem x.)
We have shown that an arbitrary instance x of VC can be solved using a polynomial number of computational steps, plus a polynomial number of calls to an "oracle" that solves IS. In other words we have proved that VC ≤P IS.
Additionally, a similar kind of proof will show that IS ≤P VC. Thus, one can prove that VC ≡P IS.