Interpretation of co-NPCompleteness?
As far as I understand your problem statement is as following:


  Given a Problem $A$ that has an answer $true$ if and only if both (some) conditions 1 or 2 are $false$.


We have a decision problem $A(w) = \overline{B(w)} \wedge \overline{C(w)}$, where $B$ and $C$ correspond to the conditions 1 and 2 respectively.


  Whether condition 2 is $true$ can be tested with certainty in polynomial time. 


$C \in P$.


  Checking if condition 1 is $true$ on $A$ is an NPComplete problem (if we consider all possible instances of $A$).


$B \in NP$ and $B$ is $NP$-hard.

(Please, correct me if I misunderstood the problem statement)

So, let's rewrite the statement using the language notation: $L_B \in NP$, $L_C \in P$, and $L_A = \overline{L}_B \cap \overline{L}_C$. 

By definition $L_A \in coNP$ if $\overline{L}_A \in NP$. But $\overline{L}_A = L_B \cup L_C$ (De Morgan law). Since we are given that $L_B \in NP$ and $L_C \in P$, $L_B \cup L_C$ is in $NP$ as well. So, $L_A$ is in $coNP$.

Now, take $\overline{L_C} = \emptyset$ which is clearly is in $P$, and so is $L_C = \Sigma^*$. For any $NP$-complete language $L_B$, $L_B \cup L_C = \Sigma^*$. But $\Sigma^*$ is not $coNP$-complete.