依賴選擇公理

${\displaystyle (\,\Longleftarrow \,)}$

${\displaystyle R}$

${\displaystyle X}$

${\displaystyle X}$

${\displaystyle T}$

${\displaystyle R}$

${\displaystyle T}$

${\displaystyle R}$

${\displaystyle 0\leq k<n.}$

${\displaystyle x_{k}R\,x_{k+1}}$

${\displaystyle \langle x_{0},\dots ,x_{n}\rangle \in T}$

${\displaystyle R}$

${\displaystyle T}$

${\displaystyle \omega }$

${\displaystyle T}$

${\displaystyle \langle x_{0},\dots ,x_{n},\dots \rangle .}$

${\displaystyle n\geq 0\!:}$

${\displaystyle \langle x_{0},\dots ,x_{n},x_{n+1}\rangle \in T,}$

${\displaystyle x_{n}R\,x_{n+1}.}$

${\displaystyle {\mathsf {DC}}}$

${\displaystyle (\,\Longrightarrow \,)}$

設${\displaystyle T}$是一棵位於${\displaystyle X}$上具有${\displaystyle \omega }$層的剪枝過的樹，那麼此處的策略是定義${\displaystyle T}$上的二元關係${\displaystyle R}$，而這關係使得${\displaystyle {\mathsf {DC}}}$導出${\displaystyle t_{n}=\langle x_{0},\dots ,x_{f(n)}\rangle }$這樣的序列，而在這序列中，${\displaystyle t_{n}R\,t_{n+1}}$且${\displaystyle f(n)}$是一個嚴格遞增函數；而在這種狀況下，無窮序列${\displaystyle \langle x_{0},\dots ,x_{k},\dots \rangle }$是一個分支。（要證明這點，只需要對${\displaystyle f(n)=m+n.}$進行證明）我們先定義「若${\displaystyle u}$是${\displaystyle v}$的始序列（initial subsequence），且${\displaystyle \operatorname {length} (u)>0,\,}$且${\displaystyle \operatorname {length} (v)=}$ ${\displaystyle \operatorname {length} (u)+1.}$，則${\displaystyle u\,R\,v}$」開始，由於${\displaystyle T}$是一棵具有${\displaystyle \omega }$層的剪枝過的樹枝故，所以${\displaystyle R}$是個完整關係；因此${\displaystyle {\mathsf {DC}}}$蘊含說存在有無限序列${\displaystyle t_{n}}$使得${\displaystyle t_{n}\,R\,t_{n+1}}$，因此對於一些${\displaystyle m\geq 0.}$而言，${\displaystyle t_{0}=\langle x_{0},\dots ,x_{m}\rangle }$。設${\displaystyle x_{m+n}}$${\displaystyle t_{n}.}$的最終元素，那麼${\displaystyle t_{n}=\langle x_{0},\dots ,x_{m},\dots ,x_{m+n}\rangle .}$。對於所有的${\displaystyle k\geq 0,}$而言${\displaystyle \langle x_{0},\dots ,x_{k}\rangle }$這序列屬於${\displaystyle T}$。由於這是${\displaystyle t_{0}\,(k\leq m)}$的的始序列，或者是一個${\displaystyle t_{n}\,(k\geq m)}$之故，因此${\displaystyle \langle x_{0},\dots ,x_{k},\dots \rangle }$是一個分支。