Figure 8: Internal structure of the nuclides of helium, as proposed by the Cordus theory. In Figure (a) we have the following. (1) Prohibited: two cisphasic protons do not give complete HEDs. (2) Cisphasic bonds are possible in principle but are nonviable without at least one neutron per cube. In Figure (b) we have the following: (1) Open polymer: protons are stable with an externally open end (but neutrons are not), which applies to 2He1 (shown here) and 1H0. (2) All these are cisphasic joints; this is proposed as the reason why the assembly is stable. (3) The two proton ends are open, because the structure is simple enough to be fully positionally determined. In Figure (c) we have the following. (1) Lamellar plate structure: four nucleons in a square, for 2He2. This is the nominal representation: the actual shape expected to be equal to strain on all members; that is, the square is expected to be twisted. (2) Cisphasic joints throughout. These are stable. (3) Structure is likely bent out of plane too. In Figure (d) we have the following. (1) Incomplete cube is nonviable. (2) Additional neutrons assembled with transphasic bonds into an extended loop. The two protons are unaffected. (3) Strong cisphasic joints between proton and neutron. (4) Vulnerable transphasic joints between neutrons. In Figure (e) we have the following. (1) Additional neutron added to loop. (2) Cube is now full on all available sides. (3) Symmetrical lamellar structures. In Figure (f) we have the following. (1) Extremely poor filling of the cube makes for poor stability. (2) No bridge neutron possible. (3) Asymmetrical structure. (4) Lack of proton in this cube. In Figure (g) we have the following. (1) Structure comprises 2-linear incomplete symmetrical. (2) Balanced occupancy of cubes. In Figure (h) we have the following. (1) This assembly requires a bridge neutron, but there are insufficient protons to allow this. A proton is needed here to create a viable joint. (2) Lack of symmetry. (3) Not a viable joint. In Figure (i) we have the following. (1) Both cubes are full. (2) The long neutron chain results in poor viability. (3) The series stops here because an additional cube would require the availability of another proton.