A Solution

1. First determine and name the various propositions.
   
   • N: The narcoleptic nutria is newly nidicolous

   • M: The mouflon is massively mendacious

   • L: The lapidarian is liltingly lyrical

2. We want to prove: $N \vee M$

3. We are given:
   • $L \wedge \neg M \rightarrow N$

   • $N \rightarrow \neg M$

   • $L$

4. Sample Proof:
   
   1. $L \wedge \neg M \rightarrow N$, $N \rightarrow \neg M$, $L \Rightarrow N \vee M$

   2. $\neg (L \wedge \neg M) \vee N$, $\neg N \vee \neg M$, $L \Rightarrow N \vee M$ (R5)

   3. $\neg L \vee M \vee N$, $\neg N \vee \neg M$, $L \Rightarrow N \vee M$ (DeMorgan's)

   4. $\neg L \vee M \vee N$, $\neg N \vee \neg M$, $L \Rightarrow N$, $M$ (R2)

   5. Using R3 splits this sequent into 3 new sequents:
      $\neg L$, $\neg N \vee \neg M$, $L \Rightarrow N$, $M$ (R3 #1)
      $\neg N \vee \neg M$, $L \Rightarrow N$, $M$, $L$ (R1)
      This is a valid axiom by R6

   6. $M$, $\neg N \vee \neg M$, $L \Rightarrow N$, $M$ (R3 #2)
      This is a valid axiom by R6

   7. $N$, $\neg N \vee \neg M$, $L \Rightarrow N$, $M$ (R3 #3)
      This is a valid axiom by R6

All the sequents are resolved into valid axioms, so the proof by Wang's is completed.