FALSE: the context-free pumping lemma characterizes the context-free languages

We verify the ordinary pumping condition with pumping length $p:=3$. Let $z=s^{r}q\in L$ have length at least $3$, with $q\in \{a,b,c{\}}^{\ast }$. If $r=0$, pump the first letter of $q$. If $r=1$, then $q\in K$, and pump the unique initial $s$. If $r\ge 2$ and $q$ is nonempty, pump its first letter. In all three cases take the pumped letter as $v$ and take $w=x=\epsilon$; pumping preserves membership because it either keeps $r\mathrm{\ne }1$ or changes the single initial $s$ to $s^i$, which belongs to the second part of $L$ for every $i\mathrm{\ne }1$.

Suppose for contradiction that $L$ were context free. Let $R:=\{s{a}^i{b}^j{c}^k:i,j,k\ge 0\}$, which is regular. Choose a PDA for $L$ using [L3] and a DFA for $R$. The product control consisting of the PDA state and DFA state, with the same stack operation as the PDA and with epsilon moves leaving the DFA state fixed, is a PDA for $L\cap R$. Hence $L\cap R$ is context free.

It remains to treat $z=s^r$ with $r\ge 3$. For $r=3$, take $v=s^3$; for $r\ge 4$, take $v=s^2$. Put $u=w=x=\epsilon$ and let $y$ be the remaining suffix. Pumping produces respectively $s^{3i}$ or $s^{r+2(i-1)}$, whose exponent is never $1$. Thus $\mathrm{\mid }vx\mathrm{\mid }\ge 1$, $\mathrm{\mid }vwx\mathrm{\mid }\le 3$, and every pumped word lies in $L$.

By definition, $L\cap R=sK$. Apply the homomorphism that deletes the initial symbol $s$ and fixes $a,b,c$. Closure under homomorphism from [L3] would make $K$ context free, contradicting [L2]. Therefore $L$ is not context free.

Steps 1.1 and 2.1 show that $L$ satisfies the ordinary pumping condition, while step 2.2 shows that it is not context free. Thus [A1] is false.