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To prove that a problem $\Pi_2$ is NP-hard one has to:

  1. select a known NP-hard problem $\Pi_1$;
  2. from an arbitrary instance of $\Pi_1$, create an instance of $\Pi_2$ in polynomial-time; and
  3. show that solve $\Pi_1$ with the given instance $\iff$ solve $\Pi_2$ with the created instance.

Now, my question is the following.

Suppose now I did 1., 2. and only solve $\Pi_1\Rightarrow$ solve $\Pi_2$ but when I am trying to show the inverse, i.e., solve $\Pi_2\Rightarrow$ solve $\Pi_1$, the instance of $\Pi_1$ is changed a little. Is this proof correct?

More precisely, let $I_1=(n, a_1,\ldots,a_n,b_1,\ldots,b_n,\alpha,\beta,k)$ be an arbitrary instance of $\Pi_1$. The created instance (in polynomial-time) of $\Pi_2$ is $I_2=(n, a_1,\ldots,a_n,b_1,\ldots,b_n,\Delta,k)$. Now when I tried to show solve $\Pi_2$ with $I_2\Rightarrow$ solve $\Pi_1$ with $I_1$, I found another instance of $\Pi_1$, $I_1'=(n, a_1,\ldots,a_n,b_1,\ldots,b_n,a,b,k)$ where $a\neq \alpha$ and $b\neq\beta$. Hence I did not solve $\Pi_1$ with $I_1$ but with $I_1'$.

My guess is that the proof still correct, right?

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No, you may not modify $I_1$. By the time you get to Step 3, everything is determined and you might not change your instances. Say $f$ is the polytime transformation from Step 2. Then you need to show

$I_1$ is a yes-instance of $\Pi_1$ if and only if $f(I_1) = I_2$ is a yes-instance of $\Pi_2$.

If after arriving at Step 3 you notice that your $I_2$ instance does not allow you to prove the statement, then you might want to change $f$ in your Step 2, and redo Step 3.

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