Will you look at my answer to the following homework and tell me where I made a mistake? Thanks for your help!
Let $\mathbb{P} = (P, \leq)$ be a forcing poset and let $p,q \in P$ denote two different conditions. We define the following $\mathbb{P}$-names:
\begin{align} &\underset{\widetilde{\hphantom{x}}}{x} := \{ \langle \emptyset , p \rangle\} \\ &\underset{\widetilde{\hphantom{y}}}{y} := \{ \langle \emptyset , q \rangle, \langle \underset{\widetilde{\hphantom{x}}}{x}, q \rangle\}\\ &\underset{\widetilde{\hphantom{z}}}{z} := \{ \langle \underset{\widetilde{\hphantom{y}}}{y} , p \rangle, \langle \underset{\widetilde{\hphantom{x}}}{x} , q \rangle, \langle \underset{\widetilde{\hphantom{x}}}{x} , p \rangle\}\\ &\underset{\widetilde{\hphantom{u}}}{u} := \{ \langle \underset{\widetilde{\hphantom{z}}}{z} , p\rangle, \langle \underset{\dot{\hphantom{q}}}{q} , p\rangle, \langle \underset{\dot{\hphantom{p}}}{p} , q\rangle \}\\ &\underset{\widetilde{\hphantom{v}}}{v} := \{ \langle \underset{\widetilde{\hphantom{u}}}{u} , q\rangle, \langle \underset{\dot{\hphantom{p}}}{p} , p\rangle, \langle \underset{\widetilde{\hphantom{z}}}{z} , p \rangle \} \end{align}
Determine $ \underset{\widetilde{\hphantom{v}}}{v}[\{ p \}], \underset{\widetilde{\hphantom{v}}}{v}[\{ q \}], \underset{\widetilde{\hphantom{v}}}{v}[\{ p, q \}]$
My answer:
\underset{\widetilde{\hphantom{v}}}{v}[\{ p \}] = \{ \underset{\dot{\hphantom{p}}}{p}[\{ p \} ], \underset{\widetilde{\hphantom{z}}}{z}[\{ p \}] \} = \{ \underset{\dot{\hphantom{p}}}{p}[\{ p \} ], \{ \emptyset, \{ \emptyset \} \} \}
Because \begin{align} &\underset{\widetilde{\hphantom{x}}}{x}[\{ p \}] = \{ \emptyset \}\\ &\underset{\widetilde{\hphantom{y}}}{y}[\{ p \}] = \emptyset \\ & \underset{\widetilde{\hphantom{z}}}{z}[\{ p \}] = \{ \underset{\widetilde{\hphantom{y}}}{y}[\{ p \}], \underset{\widetilde{\hphantom{x}}}{x}[\{ p \}]\} = \{ \emptyset, \{ \emptyset \} \} \end{align}
The next one is
\underset{\widetilde{\hphantom{v}}}{v}[\{ q \}] = \{ \underset{\widetilde{\hphantom{u}}}{u}[\{ q \}] \} = \{ \emptyset \}
Because
\underset{\widetilde{\hphantom{u}}}{u}[\{ q \}] = \{ \underset{\dot{\hphantom{p}}}{p}[\{ q \} ] \}= \emptyset
And the last one:
\begin{align*} &\underset{\widetilde{\hphantom{v}}}{v}[\{ p, q \}] = \\ &\{ \underset{\widetilde{\hphantom{u}}}{u}[\{ p, q \}], \underset{\dot{\hphantom{p}}}{p}[\{ p, q \}], \underset{\widetilde{\hphantom{z}}}{z}[\{ p, q \}] \} = \\ &\{ \{ \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}, \underset{\dot{\hphantom{q}}}{q}[\{ p, q \} ] , \underset{\dot{\hphantom{p}}}{p}[\{ p, q \} ] \}, \underset{\dot{\hphantom{p}}}{p}[\{ p, q \}], \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \} \} \end{align*}
Because
\begin{align} &\underset{\widetilde{\hphantom{u}}}{u}[\{ p, q \}] = \{ \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \}, \underset{\dot{\hphantom{q}}}{q}[\{ p, q \} ] , \underset{\dot{\hphantom{p}}}{p}[\{ p, q \} ] \} \\ &\underset{\widetilde{\hphantom{z}}}{z}[\{ p, q \}] = \{ \{ \emptyset \}, \{ \emptyset, \{ \emptyset \} \} \} \end{align}