As peat, the order 4 pentagonal tiling (o4o5x), has radius sqrt[-(1+sqrt(5)]/2 = 0.899454 i, it clearly is not hemi-choral within either honeycomb x3o5x *b4o and x x5o4o, while those latter two indeed have the same radius. Thence that bollocell o4o5x still can be blended out by an according overlay.

Since the vertex figure of the former is a square frustrum xf4oo&#q and that of the latter is a square pyramid of4oo&#q, the one in here becomes the tegum sum of 2 orthogonally arranged, axially-intersecting q-scaled regular pentagons. Thus it clearly is still circumscribable, just as generally required for vertex figures of uniform polytopes, and thus moreover is convex. This in turn shows, that the two laminates of srids and octs (from the first honeycomb) and of pips (from the second honeycomb) attach as disjunct laminates (and won't pleat back).

Incidence matrix according to Dynkin symbol

xØx3o5x *c4o   (N → ∞)

. . . .    . | 30N |   1   4   4 |   4   4   4   4 |   4  4  1
-------------+-----+-------------+-----------------+----------
x . . .    . |   2 | 15N   *   * |   4   0   0   0 |   4  0  0
. x . .    . |   2 |   * 60N   * |   0   2   1   0 |   0  2  1
. . . x    . |   2 |   *   * 60N |   1   0   1   2 |   2  2  0
-------------+-----+-------------+-----------------+----------
x . . x    . |   4 |   2   0   2 | 30N   *   *   * |   2  0  0
. x3o .    . |   3 |   0   3   0 |   * 40N   *   * |   0  1  1
. x . x    . |   4 |   0   2   2 |   *   * 30N   * |   0  2  0
. . o5x    . |   5 |   0   0   5 |   *   *   * 24N |   1  1  0
-------------+-----+-------------+-----------------+----------
x . o5x    . ♦  10 |   5   0  10 |   5   0   0   2 | 12N  *  *
. x3o5x    . ♦  60 |   0  60  60 |   0  20  30  12 |   * 2N  *
. x3o . *c4o ♦   6 |   0  12   0 |   0   8   0   0 |   *  * 5N