Let $\varphi$ be any propositional formula in negation normal form (NNF). A literal $\ell$ is pure in a formula $\varphi$, if the complement of $\ell$, $\ell^c$, does not occur in $\varphi$, where $\ell^c$ is $b$ if $\ell$ is $\neg b$ and $\ell^c$ is $\neg b$ if $\ell$ is $b$.
I have to prove that: If $\varphi$ contains only pure literals, then $\varphi$ is satisfiable.
I decided to prove this by structural induction over the definition of well formed NNF formulas:
Let $P(\varphi)$ denote the statement that $\varphi$ is satisfiable.
Base Case: $\varphi = a$. Then $P(\varphi)$ holds trivially because a formula which only consists of a single literal is trivially satisfiable.
Induction Hyp.: Let $\varphi$ be a formula in NNF with $n$ literals and assume $P(\varphi)$ holds.
Step Cases: $\psi := \varphi \land \ell$: By induction hyp, $\varphi$ does only contain pure litarals and $P(\varphi)$ holds. If we add another pure literal as conjuct to $\varphi$, $\psi$ stays satisfiable, since $\ell$ is a pure literal (i.e. $\ell^c$ does not occur in $\varphi$).
I believe that until here my proof structure is correct. However, can anybody help me out with the step cases? I know that I need to show it for formulas of the kind $\neg \psi$,$\psi \land \phi$, and $\psi \lor \phi$ Can anybody explain how to proceed with one of the step cases?