Ping pong for two loxodromics

Fix integers $p,q\ge N$ and set $A=g^p$, $B=h^q$. For letters $s\in \{A,A^{-1},B,B^{-1}\}$ write $D_s$ for the corresponding one of the four domains. The assumed inclusions say precisely $s(Z\setminus D_{s^{-1}})\subseteq D_s$. Every $D_s$ is nonempty, and distinct letter domains are disjoint.

Let $w=s_1\cdots s_l$ be any nonempty reduced word in these formal letters. Choose a letter $t$ different from both $s_1$ and $s_l^{-1}$; among four letters at most two are excluded. Choose $z\in D_t$. Apply the word to $z$ from right to left. Since $t\mathrm{\ne }{s}_l^{-1}$, the first application sends $z$ into $D_{s_l}$. Inductively, a point in $D_{s_{j+1}}$ lies outside $D_{s_j^{-1}}$ because reducedness means $s_{j+1}\mathrm{\ne }{s}_j^{-1}$ and the domains are disjoint. Thus applying $s_j$ sends it into $D_{s_j}$. Finally $wz\in D_{s_1}$, disjoint from the initial domain $D_t$. Therefore $wz\mathrm{\ne }z$, so the group element represented by $w$ is not the identity. This works also for $l=1$.

By F1 there is a homomorphism from the reduced-word free group on two formal generators to the ambient group sending them to $A,B$. Its image is $⟨A,B⟩$: every product of $A,B$ and their inverses is an image, and such products form that subgroup. Step 2.1 shows its kernel has no nonempty reduced word; by F1 the empty word is the only remaining element. The map is therefore injective and identifies its image with the rank-two free group. For instance the reduced commutator word $AB{A}^{-1}{B}^{-1}$ is nonidentity by exactly the same domain test, rather than by an assumed independence theorem. Only one point from one specified nonempty domain was needed for each word, so AC is not used.