Pumping Lemma Confusing

Question

Note: $\lambda $ means empty in this case.

If $v\neq\lambda$, then $|v|\ge1$. But then, the 2nd condition allows $uv^0w$ where $v^0=\lambda$ and $|v|=0$???

Shouldn't the 2nd condition only allow $i\ge1$ to maintain the 1st condition?

Answer 1

$v^0 = \lambda$ says absolutely nothing about the length of $v$. It is true for all words by definition (the concatenation of zero copies of $v$ is always empty).

$v\not = \lambda$ is a condition for selecting a factor $v$. It does not carry over to possible powers of $v$, which are in general different from $v$. For natural numbers, the fact that $5^0 =1$ does not make $5\not= 1$ untrue either.