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Indeed as in the first step of the process there are two types of tile in building ''P''<sub>''n''</sub> from ''P''<sub>''n'' – 1</sub>, those attached to an edge of ''P''<sub>''n'' – 1</sub> and those attached to a single vertex. Similarly there are two types of vertex, one in which two new tiles meet and those in which three tiles meet. So provided that no tiles overlap, the previous argument shows that angles at vertices do not exceed {{pi}} and hence that ''P''<sub>''n''</sub> is a convex polygon.
It therefore has to be verified that in constructing ''P''<sub>''n''</sub> from ''P''<sub>''n'' − 1</sub>, the new triangles do not overlap with ''P''<sub>''n'' − 1</sub> except as already described and do not overlap with each other except as already described.<ref> {{harvnb|Caratheodory|1954|pages=178−180}}</ref> By convexity, the polygon ''P''<sub>''n'' − 1</sub> is the intersection of the convex half-spaces defined by full circular arcs defining its boundary.
Finally it remains to prove that the tiling formed by the union of the triangles covers the whole of the upper half plane. Any point ''z'' covered by the tiling lies in a polygon ''P''<sub>''n''</sub> and hence a polygon ''P''<sub>''n'' +1 </sub>. It therefore lies in a copy of the original triangle Δ as well as a copy of ''P''<sub>2</sub> entirely contained in ''P''<sub>''n'' +1 </sub>. The hyperbolic distance between Δ and the exterior of ''P''<sub>2</sub> is equal to ''r'' > 0. Thus the hyperbolic distance between ''z'' and points not coverered by the tiling is at least ''r''. Since this applies to all points in the tiling, the set covered by the tiling is closed. On the other hand the tiling is open since it coincides with the union of the interiors of the polygons ''P''<sub>''n''</sub>. By connectivity, the tessellation must cover the whole of the upper half plane.
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