Can somebody show that BPTT, version 2 is deduced from BPTT, version 1
[BPTT, version 1] Let $h$ be a fixed point free orientation preserving homeomorphism of $\mathbb{R}^2$ Every point $m\in\mathbb{R}^2$ belongs to a properly embedded topological line $\bigtriangleup $ which separates $h^{−1}(\bigtriangleup )$ and $h(\bigtriangleup)$.
[BPTT, version 2] Let $h$ be a fixed point free orientation preserving homeomorphism of $\mathbb{R}^2$. For every point $m\in\mathbb{R}^2$ there exists a topological embedding $\varphi : \mathbb{R}^2\to\mathbb{R}^2$ such that
$m\in\varphi(\mathbb{R}^2)$,
$\forall\,\,\, x\in\mathbb{R}\,\,\,\,\varphi(\{x\}\times\mathbb{R})$ is a closed subset of $\mathbb{R}^2$,
$h\circ\varphi = \varphi\circ\tau$ where $\tau $ is the translation $\tau (x, y) = (x + 1, y)$.
For going from version 2 to version 1, try letting $\Delta$ be the line $\phi({m}\times \mathbb{R})$?
For going the other direction, it's a lot harder. Try defining $\phi$ first by sending the $\Delta$ containing some $m$ to the vertical $y$-axis, then sending $h^n(\Delta))$ to the line $x=n$ for all integers $n$ (defining the map to satisfy property 3). Then fill in the vertical strip from $x=0$ to 1 by an arbitrary homeomorphism with the strip between $\Delta$ and $h(\Delta)$, and then define the rest of the map by property 3.