that the crystal $\M(p_1,p_2;r;\infty)$ is isomorphic to $\M(\infty)$ as a $\uq(G_2)$-crystal.

First, we define a canonical map $\phi:\M(\infty)\rightarrow\M(p_1,p_2;r;\infty)$ by setting $$\begin{aligned}
\phi(M)=& X_1(r-1)^{(p_1+p_2,-b^1_2-b^1_3-b^1_0-
                      b^1_{\bar3}-b^1_{\bar2}-b^1_{\bar1})}
          X_2(r-1)^{(0,b_2^1)}
          X_3(r-1)^{(0,b_3^1)}\\
        & \cdot
          X_0(r-1)^{(0,b_0^1)}
          X_{\bar3}(r-1)^{(0,b_{\bar3}^1)}
          X_{\bar2}(r-1)^{(0,b_{\bar2}^1)}
          X_{\bar1}(r-1)^{(0,b_{\bar1}^1)}\\
        & \cdot
          X_2(r-2)^{(p_2,-b_3^2)}
          X_3(r-2)^{(0,b_3^2)}
\end{aligned}$$ for $M$ of the form . Note that the monomial $M_{\infty}$ of $\M(\infty)$ is mapped onto the vector $M_{(p_1,p_2;r;\infty)}$. It is obvious that this map $\phi$ is well-defined and that it is actually bijective.

The outputs of the functions $\operatorname{wt}$, $\vphi_i$, and $\veps_i$, defined in , , and do not depend on $r$ or on any fixed positive integers $p_i$. Using Lemma \[anaysiskashi2\] and its counterpart for $\M(p_1,p_2;r;\infty)$, we can easily show that the map $\phi$ commutes with the Kashiwara operators. Hence, the set $\M(p_1,p_2;r;\infty)$ forms a subcrystal of $\Me$ isomorphic to $\M(\infty)$ as a $\uq(G_2)$-crystal.

The monomial of  is the element of $\M(p_1,p_2;r;\infty)$ corresponding to the monomial  of $\M(\infty)$ under the natural isomorphism $\phi$ mentioned in the proof of Proposition \[cor4\].

\[wkwmd3\] Since $\vepst_i(M_{\infty})=(0,0)$ for all $i\in I$, we have $\eit(M_{\infty})=0$ for all $i\in I$. And $\operatorname{\tilde{wt}}(M_{\infty})=\sum_{i\in I} (1,0)\La_i$, so we have $\operatorname{wt}(M_{\infty})$ $=0$. Note that $M_{\infty}$ satisfies the condition for a monomial $M$ mentioned in the conjecture introduced in Section \[Nakajima 