Let $n \in \mathbb { N }$ and $\gamma = \left( \gamma _ { 1 } , \ldots , \gamma _ { 2 n + 2 } \right)$ be a Dyck path of length $2 n + 2$. Let $r = \max \left\{ i \in \llbracket 0 , n \rrbracket \mid s _ { \gamma } ( 2 i ) = 0 \right\}$. We assume $0 < r < n$ and we consider the paths $\alpha = \left( \gamma _ { 1 } , \ldots , \gamma _ { 2 r } \right)$ and $\beta = \left( \gamma _ { 2 r + 2 } , \ldots , \gamma _ { 2 n + 1 } \right)$. Justify using a figure that $\gamma _ { 2 r + 1 } = 1 , \gamma _ { 2 n + 2 } = - 1$ and that $\alpha$ and $\beta$ are Dyck paths.
Let $n \in \mathbb { N }$ and $\gamma = \left( \gamma _ { 1 } , \ldots , \gamma _ { 2 n + 2 } \right)$ be a Dyck path of length $2 n + 2$. Let $r = \max \left\{ i \in \llbracket 0 , n \rrbracket \mid s _ { \gamma } ( 2 i ) = 0 \right\}$. We assume $0 < r < n$ and we consider the paths $\alpha = \left( \gamma _ { 1 } , \ldots , \gamma _ { 2 r } \right)$ and $\beta = \left( \gamma _ { 2 r + 2 } , \ldots , \gamma _ { 2 n + 1 } \right)$. Justify using a figure that $\gamma _ { 2 r + 1 } = 1 , \gamma _ { 2 n + 2 } = - 1$ and that $\alpha$ and $\beta$ are Dyck paths.