Let $X=(X_t)_{t \ge 0}$ be a strong Markov process, and let $G_1,\, G_2$, and $G_3$ be continuous functions satisfying $G_1 \le G_3 \le G_2$ and $\mathsf{E}_x\sup_t \vert G_i(X_t) \vert < \infty$ for $i=1,2,3$. Consider the optimal stopping game where the sup-player chooses a stopping time $\tau$ to maximize, and the inf-player chooses a stopping time $\sigma$ to minimize, the expected payoff $\mathsf{M}_x(\tau,\sigma) = \mathsf{E}_x [G_1(X_\tau)\;\! I(\tau\! <\! \sigma) + G_2(X_\sigma)\;\! I(\sigma\! <\! \tau) + G_3(X_\tau)\;\! I(\tau\! =\! \sigma)],$ where $X_0=x$ under $\mathsf{P}_{\!x}$. Define the upper value and the lower value of the game by $V^*(x) = \inf_\sigma \sup_\tau \mathsf{M}_x(\tau,\sigma)~{\rm and}~ V_*(x) = \sup_\tau \inf_\sigma \mathsf{M}_x(\tau,\sigma),$ respectively, where the horizon T (the upper bound for $\tau$ and $\sigma$ above) may be either finite or infinite (it is assumed that $G_1(X_T)=G_2(X_T)$ if T is finite and $\liminf_{t \rightarrow \infty} G_2(X_t) \le \limsup_{t \rightarrow \infty} G_1(X_t)$ if T is infinite). If X is right-continuous, then the Stackelberg equilibrium holds, in the sense that $V^*(x)=V_*(x)$ for all x with $V:=V^*=V_*$ defining a measurable function. If X is right-continuous and left-continuous over stopping times (quasi-left-continuous), then the Nash equilibrium holds, in the sense that there exist stopping times $\tau_*$ and $\sigma_*$ such that $\mathsf{M}_x(\tau,\sigma_*) \le \mathsf{M}_x(\tau_*,\sigma_*) \le \mathsf{M}_x(\tau_*,\sigma)$ for all stopping times $\tau$ and $\sigma$, implying also that $V(x)=\mathsf{M}_x(\tau_*,\sigma_*)$ for all x. Further properties of the value function V and the optimal stopping times $\tau_*$ and $\sigma_*$ are exhibited in the proof.