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How about the Proper Game formulation?

$(P, \leq, 1)$ is proper iff

$\exists \Sigma \ \forall \pi \ \forall p \in P :$

• IF $\forall x \in \pi [x$ is an ordered pair $(x_1, x_2)$, $x_1$ is natural, and $1 \Vdash _P\ (x_2$ is an ordinal$)]$
• THEN $\Sigma (p, \pi)$ is an ordered pair $(q, \sigma)$ such that:
1. $\forall y \in \sigma\ [y$ is an ordered pair $(y_1,y_2)$, $y_1$ is natural, and $y_2$ is an ordinal$]$
2. $q \in P\$ is such that:
i. $q \leq p$
ii. $\forall x \in \pi\ [q\ \Vdash _P\ \left ( \exists y \in \sigma \right )\left (x_2 = y_2\right )]$

In other words this is saying there's a strategy $\Sigma$ for player II such that for any play from player I, consisting of a condition $p$ and a (partial) $\omega$-sequence of $P$-names for ordinals $\pi$, $\Sigma (p, \pi)$ produces a condition $q$ extending $p$, and a (partial) $\omega$-sequence of ordinals $\sigma$ such that $\forall n \in \mathrm{dom} (\pi),\ q \Vdash \exists k \in \mathrm{dom} (\sigma) (\pi (n) = \sigma (k))$.

1

How about the Proper Game formulation?

$(P, \leq, 1)$ is proper iff

$\exists \Sigma \ \forall \pi \ \forall p \in P :$

• IF $\forall x \in \pi [x$ is an ordered pair $(x_1, x_2)$, $x_1$ is natural, and $1 \Vdash _P\ (x_2$ is an ordinal$)]$
• THEN $\Sigma (p, \pi)$ is an ordered pair $(q, \sigma)$ such that:
1. $\forall y \in \sigma\ [y$ is an ordered pair $(y_1,y_2)$, $y_1$ is natural, and $y_2$ is an ordinal$]$
2. $q \in P\$ is such that:
i. $q \leq p$
ii. $\forall x \in \pi\ [q\ \Vdash _P\ \left ( \exists y \in \sigma \right )\left (x_2 = y_2\right )]$

In other words this is saying there's a strategy $\Sigma$ for player II such that for any play from player I, consisting of a condition $p$ and a (partial) $\omega$-sequence of $P$-names $\pi$, $\Sigma (p, \pi)$ produces a condition $q$ extending $p$, and a (partial) $\omega$-sequence of ordinals $\sigma$ such that $\forall n \in \mathrm{dom} (\pi),\ q \Vdash \exists k \in \mathrm{dom} (\sigma) (\pi (n) = \sigma (k))$.