$A$ is NP-complete.
$B$ is P.
$A \cap B = \emptyset $
$A \cup B \neq \sum^{*}$
Prove that $A \cup B $ is NP-complete.
How can I prove this ? I think if anything can be P-reducible to A then it can also be reducible to $A \cup B$, but don't know how to formally say this.
Thanks.
Hint Just use the usual argument. Take a procedure that decides if something is in $A \cup B$ and turn this, with a polynomial-time transformation, into a procedure that decides if something is in $A$. For this you need to use another procedure that decides, in polynomial-time, if something is in $B$.
Concretely Let's say you have a procedure $P_{AB}$ that decides if something is in $A \cup B$, that is $P_{AB}(\omega) = \text{true}$ if $\omega \in A \cup B$ and $P_{AB}(\omega) = \text{false}$ if $\omega \not\in A \cup B$. Similarly, let's say $P_B$ is a polynomial-time procedure that decides if something is in $B$, that is $P_B(\omega) = \text{true}$ if $\omega \in B$ and $P_B(\omega) = \text{false}$ if $\omega \not\in B$.
Now we have to build a procedure $P_A(\omega)$ that decides if something is in $A$. Ofcourse, first it will call $P_{AB}(\omega)$ and $P_B(\omega)$. Now, when is $\omega \in A$ and when is $\omega \not\in A$. Recall at this point that $A \cap B = \emptyset$.