Resolution rule in propositional calculus I was thinking about a reverse case of validity of resolution rule and had a question about it.
Basically, let me state resolution rule first. 

Suppose $C_1$ and $C_2$ are clauses such that a literal $l$ belongs to $C_1$ and a complementary literal $l'$ belongs to $C_2$. Then the resolvant $C$ of $C_1$ and $C_2$ is $(C_1 \cup C_2) \setminus \{l, l'\}$

Now with this rule, we know that always $C_1 \models C$ and $C_2 \models C$. 
But, is it always true that $C \models C_1$ and $C \models C_2$?
 A: 
Now with this rule, we know that always $C_1 \models C$ and $C_2 \models C$. 

This isn't true.  Let $C_1 = \{p,q\}$ and $C_2 = \{\lnot q, r\}$. Then $C = \{p, r \}$.  It's not the case that $\{p,r\}$ must be true when $\{p,q\}$ is true. For instance, consider the truth assignment that makes $q$ true, and $p$ and $r$ false.  Then $C_1$ is satisfied, but $C$ is not.  Similarly, the truth assignment that makes $p$, $q$, and $r$ false satisfies $C_2$, but not $C$.  However, any truth assignment that satisfies both $C_1$ and $C_2$ also satisfies $C$.

But, is it always true that $C \models C_1$ and $C \models C_2$?

No.  For example, consider the truth assignment that makes $p$ and $q$ false, and $r$ true.  This satisfies $C$, but not $C_1$.  Similarly, the truth assignment that makes $p$ and $q$ true and $r$ false satisfies $C$, but not $C_2$.  However, since every literal in $C$ is in either $C_1$ or $C_2$, any truth assignment that satisfies $C$ must satisfy at least one of $C_1$ and $C_2$.
