Proof for the amount of states of a regular language 
Let L be a formal language. Two words $u,v \in \Sigma^*$ are called
  separable if $\exists w \in \Sigma^* : uw \in L, vw \not\in L$.
Nerode lemma: Let L be a formal language. If there are $n$ words of L that are pairwise separable, a final state machine that
  recognizes L would have at least $n$ states.

Let $L_i = \{ubv \; | \; u,v \in \{a,b\}^*, |v| = i - 1\}$. I now need to prove that a final state machine for $L_i$ contains at least $2^i$ states using the defintion and lemma above. Therefore there should be $2^i$ pairwise separable words. 
I tried this for $L_2$:

"-" means that the two words are not separable. The fields marked with "x" are ignored because of the symmetry. Therefore e.g. $\lambda$ and $b$ are separable by $a$ because $\lambda.a = a \not\in L \wedge b.a \in L$. Nevertheless instead of 4 pairwise separable items ($2^2$) I only got three: $\lambda,a$ and $ba$. One is missing.
Could you please help me to find the missing one and a possibility to prove it for all $L_i$?
Thanks in advance! 

Let $E = \Sigma^i$. Clearly $|E| = 2^i$. I claim that all pairs from $E$ are separable.
Suppose $u,v \in E$, with $u \neq v$. Let $k$ be the last index such that $[u]_k \neq [v]_k$. WLOG, we can take $[u]_k = b$. Then $u$ has the form $u = \alpha b \beta$, with $|\alpha| = k-1$, $|\beta| = i-k$. Choose $w = a^{k-1}$ (possibly empty).
Then we have $[uw]_{|uw|-i+1} = b$, but $[vw]_{|vw|-i+1} = a$, so we have that $uw \in L_i$, but $vw \notin L_i$.
