A problem with Special Relativity Theory for perdurantists

There seems to be a problem for the conjunction of Special Relativity and perdurantism.  Maybe this is a standard problem that has a standard solution? Let's say that being bent is an intrinsic property. Perdurantists of the sort I am interested in think that Socrates is bent at a time in virtue of an instantaneous temporal part of him being bent (I think the argument can be made to work with thin but not instantaneous parts, but it's a little more complicated). Therefore:

1. x is bent at t only if the temporal part of x at t is bent simpliciter.

The following also seems like something perdurantists should say:

2. x is bent simpliciter only if every temporal part of x is bent simpliciter.

Now, we need to add some premises about the interaction of Special Relativity and time.

3. There is a one-to-one correspondence between times and maximal spacelike hypersurfaces such that one exists at a time if and only if one at least partly occupies the corresponding hypersurface.

Given a time t, let H(t) be the corresponding maximal spacelike hypersurface. And if h is a maximal spacelike hypersurface, then let T(h) be the corresponding time. Write P(x,t) for the temporal part of x at t. Then:

4. P(x,t) is wholly contained within H(t) and if z is a spacetime point in H(t) and within x, then z is within P(x,t)

and, plausibly:

5. If a point within x is within a maximal spacelike hypersurface h, then P(x,T(h)) exists.

Now suppose we have Special Relativity, so we're in a Minkowski spacetime. Then:

6. For any point z in spacetime, there are three maximal spacelike hypersurfaces h₁, h₂ and h₃ whose intersection contains no points other than z.

Add this obvious premise:

7. No object wholly contained within a single spacetime point is bent simpliciter.

Finally, for a reductio, suppose:

8. x is an object that is bent at t.

Choose a point z within P(x,t) and choose three spacelike hypersurfaces h₁, h₂ and h₃ whose intersection contains z and only z (by 6). Now define the following sequence of objects, which exist by 4 and 5:

• x₁=P(x,t)
• x₂=P(x₁,T(h₁))
• x₃=P(x₂,T(h₂))
• x₄=P(x₃,T(h₃))

Observe that x₄ is wholly contained in the intersection of the three hypersurfaces h₁, h₂ and h₃, and hence:

9. x₄ is wholly at z.
10. It is not the case that x₄ is bent simpliciter.

Now:

11. x₁ is bent simpliciter. (By 1 and 8)
12. x₂ is bent simpliciter. (By 2 and 11)
13. x₃ is bent simpliciter. (By 2 and 12)
14. x₄ is bent simpliciter. (By 2 and 13)

Since 14 contradicts 10, we have a problem.  It seems the perdurantist cannot have any objects that are bent at any time in a Minkowski spacetime. This is a problem for the perdurantist. If I were a perdurantist, I'd deny 2, and maintain that an object can be bent simpliciter despite having temporal parts that are bent and temporal parts that are not bent. But I would not be comfortable with maintaining this. I would take this to increase the cost of perdurantism. What is ironic here is that it is often thought that endurantism is what has trouble with Relativity.